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Let $\vec{a} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = 4\hat{i} + 3\hat{j} - 2\hat{k}$. A vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by their cross product $\vec{a} \times \vec{b}$.
We compute the cross product as follows: $$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ 4 & 3 & -2 \end{vmatrix} $$ $$ \vec{a} \times \vec{b} = \hat{i}((-2)(-2) - (1)(3)) - \hat{j}((3)(-2) - (1)(4)) + \hat{k}((3)(3) - (-2)(4)) $$ $$ \vec{a} \times \vec{b} = \hat{i}(4 - 3) - \hat{j}(-6 - 4) + \hat{k}(9 + 8) $$ $$ \vec{a} \times \vec{b} = \hat{i}(1) - \hat{j}(-10) + \hat{k}(17) $$ $$ \vec{a} \times \vec{b} = \hat{i} + 10\hat{j} + 17\hat{k} $$
The magnitude of the vector $\vec{a} \times \vec{b}$ is: $$ |\vec{a} \times \vec{b}| = \sqrt{(1)^2 + (10)^2 + (17)^2} = \sqrt{1 + 100 + 289} = \sqrt{390} $$
The unit vector in the direction of $\vec{a} \times \vec{b}$ is: $$ \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{\hat{i} + 10\hat{j} + 17\hat{k}}{\sqrt{390}} $$
A vector of magnitude 5 in the direction of $\hat{n}$ is: $$ \vec{v} = 5\hat{n} = 5\left(\frac{\hat{i} + 10\hat{j} + 17\hat{k}}{\sqrt{390}}\right) = \frac{5}{\sqrt{390}}(\hat{i} + 10\hat{j} + 17\hat{k}) $$ Since the question asks for a vector perpendicular to both given vectors, the vector can also be in the opposite direction. Therefore, the required vector is $\pm \frac{5}{\sqrt{390}}(\hat{i} + 10\hat{j} + 17\hat{k})$.
Final Answer: $\pm \frac{5}{\sqrt{390}}(\hat{i} + 10\hat{j} + 17\hat{k})$