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

The two given planes are
$$2x + 3y + z = 1 \quad\text{and}\quad x + 3y + 2z = 2$$

Let $$\mathbf{n}_1$$ and $$\mathbf{n}_2$$ be their normal vectors:
$$\mathbf{n}_1 = (2,\,3,\,1), \qquad \mathbf{n}_2 = (1,\,3,\,2).$$

The line $$L$$ is the intersection of the two planes, so its direction vector $$\mathbf{d}$$ is perpendicular to both normals. Hence
$$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2.$$ Compute the cross product:

$$ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 3 & 1\\ 1 & 3 & 2 \end{vmatrix} = \mathbf{i}(3\cdot2 - 1\cdot3)\;-\;\mathbf{j}(2\cdot2 - 1\cdot1)\;+\;\mathbf{k}(2\cdot3 - 3\cdot1) = (3,\,-3,\,3). $$

Divide by $$3$$ to get a simpler parallel vector:
$$\mathbf{d} = (1,\,-1,\,1).$$

The positive $$x$$-axis has direction vector $$\mathbf{i} = (1,\,0,\,0).$$
If $$\alpha$$ is the angle between $$\mathbf{d}$$ and the positive $$x$$-axis, then by the definition of the dot product

$$ \cos\alpha = \frac{\lvert\mathbf{d}\cdot\mathbf{i}\rvert} {\lVert\mathbf{d}\rVert\,\lVert\mathbf{i}\rVert}. $$

Compute each term:
$$\mathbf{d}\cdot\mathbf{i} = 1\cdot1 + (-1)\cdot0 + 1\cdot0 = 1,$$
$$\lVert\mathbf{d}\rVert = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3},$$
$$\lVert\mathbf{i}\rVert = 1.$$

Therefore
$$ \cos\alpha = \frac{1}{\sqrt{3}}. $$

Option A which is: $$\frac{1}{\sqrt{3}}$$