Let $L$ be the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=2$. If $L$ makes an angle $\alpha$ with the positive $x$ -axis, then $\cos \alpha$ equals:

The line of intersection $L$ is perpendicular to the normals of both planes.Normal to Plane 1: $\overrightarrow{n_1}=2\hat{i}+3\hat{j}+\hat{k}$ Normal to Plane 2: $\overrightarrow{n_2}=\hat{i}+3\hat{j}+2\hat{k}$ The direction vector $\vec{v}$ of line $L$ is given by $\overrightarrow{n_1}\times \overrightarrow{n_2}$:
$$\vec{v}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 1\\ 1& 3& 2\end{array}\right|$$
$$\vec{v}=\hat{i}(6-3)-\hat{j}(4-1)+\hat{k}(6-3)$$
$$\vec{v}=3\hat{i}-3\hat{j}+3\hat{k}$$
The direction ratios of the line are $(3,-3,3)$ or $(1,-1,1)$.The direction cosines $(l,m,n)$ are:
$$l=\frac{1}{\sqrt{1^2+(-1{)}^2+1^2}}=\frac{1}{\sqrt{3}}$$
The angle $\alpha$ with the positive $x$ -axis is given by $\cos \alpha =l$.
$$\cos \alpha =\frac{1}{\sqrt{3}}$$