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

$\frac{1}{\sqrt{3}}$

Vectors normals to the given planes are $\vec{n_1}= 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{n_2}=\hat{i} + 3\hat{j} +2 \hat{k}$. So, the line L is parallel to 

$\vec{n} = \vec{n_1}× \vec{n_2}= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\2 & 3 & 1\\1 & 3 & 2\end{vmatrix}= 3\hat{i}-3\hat{j} + 3\hat{k}$

$⇒ cos \alpha = \frac{\vec{n}.\hat{i}}{|\vec{n}||\hat{i}|}= \frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}$