Answer: Answer: \(\mathbf{a} \times \mathbf{b} = \mathbf{i} - 10\mathbf{j} - 17\mathbf{k}\)

The area of triangle \(OAB\) is \(\frac{1}{2}\sqrt{390}\).

The volume of tetrahedron \(OABC\) is \(5\).

Let \(\mathbf{a}=(3,2,-1)\) and \(\mathbf{b}=(4,-3,2)\).

Then

\(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & -1 \\ 4 & -3 & 2 \end{vmatrix}\)

\(= \mathbf{i}(2\cdot 2-(-1)(-3)) - \mathbf{j}(3\cdot 2-(-1)\cdot 4) + \mathbf{k}(3\cdot(-3)-2\cdot 4)\)

\(= \mathbf{i}(4-3) - \mathbf{j}(6+4) + \mathbf{k}(-9-8)\)

\(= \mathbf{i} - 10\mathbf{j} - 17\mathbf{k}.\)

The magnitude of \(\mathbf{a} \times \mathbf{b}\) is the area of the parallelogram \(OAB\), so the area of triangle \(OAB\) is

\(\frac{1}{2}\left|\mathbf{a} \times \mathbf{b}\right| = \frac{1}{2}\sqrt{1^2+(-10)^2+(-17)^2} = \frac{1}{2}\sqrt{390}.\)

Now take \(\mathbf{c}=(3,-1,-1)\). A vector perpendicular to the plane \(OAB\) is \(\mathbf{a} \times \mathbf{b} = (1,-10,-17)\), so the perpendicular height from \(C\) to the plane \(OAB\) is the component of \(\mathbf{c}\) in this direction:

\(\text{height} = \frac{|\mathbf{c}\cdot(\mathbf{a} \times \mathbf{b})|}{|\mathbf{a} \times \mathbf{b}|}.\)

Compute the dot product:

\(\mathbf{c}\cdot(\mathbf{a} \times \mathbf{b}) = (3,-1,-1)\cdot(1,-10,-17) = 3+10+17 = 30.\)

Hence

\(\text{height} = \frac{30}{\sqrt{390}}.\)

Therefore the volume of tetrahedron \(OABC\) is

\(\frac{1}{3} \times \frac{1}{2}\sqrt{390} \times \frac{30}{\sqrt{390}} = 5.\)