The vector product can be evaluated as follows:

\( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 3 & 8 \\ 1 & -3 & 2 \end{vmatrix} \)
\(= \mathbf{i} \begin{vmatrix} 3 & 8 \\ -3 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 8 \\ 1 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 3 \\ 1 & -3 \end{vmatrix} \)
\(= \mathbf{i} \left(3 \times 2 - 8 \times (-3)\right) - \mathbf{j} \left(3 \times 2 - 8 \times 1\right) + \mathbf{k} \left(3 \times (-3) - 3 \times 1\right) \)
\(= 30\mathbf{i} + 2\mathbf{j} - 12\mathbf{k} \)
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The magnitude of \( \mathbf{a} \times \mathbf{b} \) is \( \sqrt{30^2 + 2^2 + (-12)^2} = 2\sqrt{262} \).

Hence a unit vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{1}{{2\sqrt{262}}} (30i + 2j - 12k) \).
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Proof:

By assumption, we have

\( \mathbf{a} \times \mathbf{b} = \mathbf{c} \times \mathbf{a} \)

\( \mathbf{a} \times \mathbf{b} - \mathbf{c} \times \mathbf{a} = 0 \)

\( \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} = 0 \)

Therefore, \( \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = 0 \)
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Since \( \mathbf{a} \neq \mathbf{0} \), it follows that either \( \mathbf{b} + \mathbf{c} = \mathbf{0} \) or the vectors \( \mathbf{a} \) and \( \mathbf{b} + \mathbf{c} \) are parallel.

Hence we must have \( \mathbf{b} = -\mathbf{c} \) or \( \mathbf{a} = k(\mathbf{b} + \mathbf{c}) \) for some \( k \in \mathbb{R} \).
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Solution:

We know that the formula to find the angle between two vectors is:

\[ \sin \theta = \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \]

Therefore, \( \theta = \sin^{-1}\left( \frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \right) \)

Now, we have to find the cross product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \):

\[ \mathbf{a} \times \mathbf{b} = \mathbf{i} \times (0) - \mathbf{j} \times (0) + \mathbf{k} \times (-6) = -6\mathbf{k} \]
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While finding the angle between two vectors, substitute the magnitude of the vector value, thus:

\[ |\mathbf{a} \times \mathbf{b}| = 6 \]

Therefore, \( \theta = \sin^{-1}\left( \frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}| |\mathbf{b}|} \right) \)

\[ \theta = \sin^{-1}\left( \frac{6}{5 \cdot 2} \right) \]

\[ \theta = \sin^{-1}\left( \frac{3}{5} \right) = 36.87^\circ \]

Hence, the angle between two vectors, \( \mathbf{a} \) and \( \mathbf{b} \) (\( \theta \)), is \( 36.87^\circ \).
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To simplify \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} + \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) \), we'll use the property that \( \mathbf{a} \times \mathbf{b} \) is perpendicular to \( \mathbf{a} \).

Given that \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0 \) because \( \mathbf{a} \times \mathbf{b} \) is perpendicular to \( \mathbf{a} \), we have:

\( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} + \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = 0 + \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) \)

Since the dot product is commutative, \( \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} \), the expression simplifies to:

\( 0 + (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} \)
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Solution:

To find a vector of magnitude 5 that is perpendicular to \( \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k} \) and \( \mathbf{b} = \mathbf{i} - 2\mathbf{j} \), we'll follow these steps:

1. Calculate the vector product \( \mathbf{c} = \mathbf{a} \times \mathbf{b} \).
2. Normalize \( \mathbf{c} \) by dividing it by its magnitude.
3. Scale \( \mathbf{c} \) to have a magnitude of 5.

1. Calculate the Vector Product \( \mathbf{c} = \mathbf{a} \times \mathbf{b} \)

Given \( \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k} \) and \( \mathbf{b} = \mathbf{i} - 2\mathbf{j} \), we compute:

\( \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -1 \\ 1 & -2 & 0 \end{vmatrix} \)

\( = (\mathbf{j} \cdot 0 - \mathbf{k} \cdot (-2), -(\mathbf{i} \cdot 0 - \mathbf{k} \cdot 1), \mathbf{i} \cdot (-2) - \mathbf{j} \cdot 1) \)

\( = (2, -1, -2) \)
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2. Normalize \( \mathbf{c} \)

The magnitude of \( \mathbf{c} \) is \( |\mathbf{c}| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{9} = 3 \).

So, \( \mathbf{c} \) normalized is \( \frac{1}{3} (2\mathbf{i} - \mathbf{j} - 2\mathbf{k}) \).

3. Scale \( \mathbf{c} \) to Have a Magnitude of 5

To make the magnitude 5, we scale \( \mathbf{c} \) by \( \frac{5}{3} \).

Therefore, the vector of magnitude 5 that is perpendicular to \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{5}{3}(2\mathbf{i} - \mathbf{j} - 2\mathbf{k}) = \frac{10}{3}\mathbf{i} - \frac{5}{3}\mathbf{j} - \frac{10}{3}\mathbf{k} \).
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To show that the area of the triangle formed by the position vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) is given by \( \frac{1}{2} | \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} | \), we can use the properties of the cross product and the formula for the area of a triangle.

Let's denote the area of the triangle formed by \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) as \( A \).

The area of a triangle formed by vectors \( \mathbf{u} \) and \( \mathbf{v} \) is half the magnitude of their cross product, i.e., \( A = \frac{1}{2} | \mathbf{u} \times \mathbf{v} | \).

Therefore, the area of the triangle formed by \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) can be written as:

$$ A = \frac{1}{2} | \mathbf{a} \times \mathbf{b} | + \frac{1}{2} | \mathbf{b} \times \mathbf{c} | + \frac{1}{2} | \mathbf{c} \times \mathbf{a} | $$

Now, using the properties of the cross product:

$$ | \mathbf{a} \times \mathbf{b} | = | -\mathbf{b} \times \mathbf{a} | $$

$$ | \mathbf{b} \times \mathbf{c} | = | -\mathbf{c} \times \mathbf{b} | $$

$$ | \mathbf{c} \times \mathbf{a} | = | -\mathbf{a} \times \mathbf{c} | $$
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We can rewrite the area \( A \) as:

$$ A = \frac{1}{2} (| \mathbf{a} \times \mathbf{b} | + | \mathbf{b} \times \mathbf{c} | + | \mathbf{c} \times \mathbf{a} |) + \frac{1}{2} (| -\mathbf{b} \times \mathbf{a} | + | -\mathbf{c} \times \mathbf{b} | + | -\mathbf{a} \times \mathbf{c} |) $$

$$ = \frac{1}{2} (| \mathbf{a} \times \mathbf{b} | + | \mathbf{b} \times \mathbf{c} | + | \mathbf{c} \times \mathbf{a} | + | \mathbf{b} \times \mathbf{a} | + | \mathbf{c} \times \mathbf{b} | + | \mathbf{a} \times \mathbf{c} |) $$

$$ = \frac{1}{2} (| \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} | + | \mathbf{b} \times \mathbf{a} + \mathbf{c} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} |) $$

$$ = \frac{1}{2} | \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} + \mathbf{b} \times \mathbf{a} + \mathbf{c} \times \mathbf{b} + \mathbf{a} \times \mathbf{c} | $$

Since \( | \mathbf{b} \times \mathbf{a} | = | \mathbf{c} \times \mathbf{b} | = | \mathbf{a} \times \mathbf{c} | \), we can rewrite the equation as:

$$ A = \frac{1}{2} | \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} | $$

Therefore, the area of the triangle formed by the position vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) is given by \( \frac{1}{2} | \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} | \).
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Solution:

To find the area of the parallelogram \(OABC\), we first need to find the area of the triangle formed by the vectors \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\). Let's calculate that first.

Given the vertices \(O(0,0,0)\), \(A(0,1,3)\), and \(B(0,2,5)\), we can calculate the vectors \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) as follows:

\[
\overrightarrow{OA} = \langle 0 - 0, 1 - 0, 3 - 0 \rangle = \langle 0, 1, 3 \rangle
\]
\[
\overrightarrow{OB} = \langle 0 - 0, 2 - 0, 5 - 0 \rangle = \langle 0, 2, 5 \rangle
\]

Now, let's find the cross product of these vectors to get the area of the triangle \(OAB\):

\[
\overrightarrow{OA} \times \overrightarrow{OB} = -\mathbf{i}
\]

The magnitude of \( \overrightarrow{OA} \times \overrightarrow{OB} \) is \( | \overrightarrow{OA} \times \overrightarrow{OB} | = 1 \). Therefore, the area of the triangle \( OAB \) is \( \frac{1}{2} \times 1 = \frac{1}{2} \).

Now, since the area of a parallelogram is twice the area of a triangle formed by its adjacent sides, the area of the parallelogram \( OABC \) is \( 2 \times \frac{1}{2} = 1 \).

So, the area of the parallelogram \( OABC \) is \( 1 \).
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