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(C) The line of intersection of two planes is perpendicular to the normal vectors of both planes. The normal vectors are $\vec{n}_1 = \hat{i} + 2\hat{

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$\left(\frac{3}{\sqrt{35}}, \frac{1}{\sqrt{35}}, \frac{-5}{\sqrt{35}}\right)$

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(C) The line of intersection of two planes is perpendicular to the normal vectors of both planes. The normal vectors are $\vec{n}_1 = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{n}_2 = 2\hat{i} - \hat{j} + \hat{k}$.The direction vector $\vec{v}$ of the line of intersection is given by the cross product $\vec{n}_1 	imes \vec{n}_2$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ 2 & -1 & 1 \end{vmatrix} = \hat{i}(2 - (-1)) - \hat{j}(1 - 2) + \hat{k}(-1 - 4) = 3\hat{i} + \hat{j} - 5\hat{k}$.The magnitude of the vector $\vec{v}$ is $|\vec{v}| = \sqrt{3^2 + 1^2 + (-5)^2} = \sqrt{9 + 1 + 25} = \sqrt{35}$.The direction cosines are obtained by dividing the components of $\vec{v}$ by its magnitude:$l = \frac{3}{\sqrt{35}}, m = \frac{1}{\sqrt{35}}, n = \frac{-5}{\sqrt{35}}$.Thus,the direction cosines are $\left(\frac{3}{\sqrt{35}}, \frac{1}{\sqrt{35}}, \frac{-5}{\sqrt{35}}ight)$.

The direction cosines of the line of intersection of the planes $x+2y+z-4=0$ and $2x-y+z-3=0$ are

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