Find all vectors of magnitude $10 \sqrt{3}$ that are perpendicular to the plane of $\hat{i}+2 \hat{j}+\hat{k}$ and $-\hat{i}+3 \hat{j}+4 \hat{k}$.

(A) Let $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = -\hat{i} + 3 \hat{j} + 4 \hat{k}$.
The vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$ is given by the cross product $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -1 & 3 & 4 \end{vmatrix} = \hat{i}(8 - 3) - \hat{j}(4 + 1) + \hat{k}(3 + 2) = 5\hat{i} - 5\hat{j} + 5\hat{k}$.
The magnitude of this vector is $|\vec{a} \times \vec{b}| = \sqrt{5^2 + (-5)^2 + 5^2} = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3}$.
The unit vector perpendicular to the plane is $\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{5\hat{i} - 5\hat{j} + 5\hat{k}}{5\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$.
Vectors of magnitude $10\sqrt{3}$ perpendicular to the plane are given by $\pm 10\sqrt{3} \times \hat{n} = \pm 10\sqrt{3} \times \frac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k}) = \pm 10(\hat{i} - \hat{j} + \hat{k})$.
Thus,the required vectors are $10\hat{i} - 10\hat{j} + 10\hat{k}$ and $-10\hat{i} + 10\hat{j} - 10\hat{k}$.