1. Key Concepts and Formulas

   • Normal Vector of a Plane: For a plane given by the equation $Ax+By+Cz=D$, the normal vector to the plane is $\vec{n} = (A, B, C)$. This vector is perpendicular to the plane.
   • Direction Vector of the Line of Intersection: The line of intersection of two planes is perpendicular to the normal vector of each plane. Therefore, its direction vector $\vec{d}$ is parallel to the cross product of the normal vectors of the two planes, i.e., $\vec{d} \propto \vec{n_1} \times \vec{n_2}$.
   • Direction Cosine with the x-axis: If a line has a direction vector $\vec{d} = (d_x, d_y, d_z)$, the cosine of the angle $\alpha$ it makes with the positive x-axis is given by $\cos \alpha = \frac{d_x}{|\vec{d}|}$.

2. Step-by-Step Solution

   Step 1: Identify the normal vectors of the given planes.

   • What we are doing: Extracting the coefficients of $x, y, z$ from the plane equations to find their normal vectors.
   • Why: The normal vectors are crucial for determining the orientation of the planes, which in turn helps find the direction of their line of intersection.
   • The equations of the given planes are:
     • Plane 1: $2x+3y+z=1$
     • Plane 2: $x+3y+2z=2$
   • From these equations, we identify the normal vectors:
     • For Plane 1: $\vec{n_1} = (2, 3, 1)$
     • For Plane 2: $\vec{n_2} = (1, 3, 2)$

   Step 2: Calculate the direction vector of the line of intersection.

   • What we are doing: Computing the cross product of the normal vectors.
   • Why: The line of intersection lies in both planes, meaning it is perpendicular to both normal vectors. The cross product of two vectors yields a vector that is perpendicular to both original vectors, thus providing the direction vector of the line of intersection.
   • Let $\vec{d}$ be the direction vector of the line of intersection $L$. We calculate $\vec{d} = \vec{n_1} \times \vec{n_2}$: $\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix}$
   • Expanding the determinant: $\vec{d} = \hat{i}((3)(2) - (1)(3)) - \hat{j}((2)(2) - (1)(1)) + \hat{k}((2)(3) - (3)(1))$ $\vec{d} = \hat{i}(6 - 3) - \hat{j}(4 - 1) + \hat{k}(6 - 3)$ $\vec{d} = 3\hat{i} - 3\hat{j} + 3\hat{k}$
   • So, a direction vector for the line $L$ is $(3, -3, 3)$.
   • Tip: We can use any non-zero scalar multiple of this vector as the direction vector. To simplify calculations, we can divide by 3: $\vec{d'} = (1, -1, 1)$

   Step 3: Determine $\cos \alpha$.

   • What we are doing: Using the direction cosine formula to find the cosine of the angle with the positive x-axis.
   • Why: The direction cosine formula directly relates the x-component of the direction vector and its magnitude to the angle with the x-axis.
   • First, calculate the magnitude of the simplified direction vector $\vec{d'} = (1, -1, 1)$: $|\vec{d'}| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$
   • The x-component of $\vec{d'}$ is $d_x = 1$.
   • Now, calculate $\cos \alpha$: $\cos \alpha = \frac{d_x}{|\vec{d'}|} = \frac{1}{\sqrt{3}}$

3. Common Mistakes & Tips

   • Cross Product Calculation: Be very careful with the signs and order of multiplication when calculating the cross product. A common error is mixing up the $\hat{j}$ component's negative sign.
   • Magnitude Calculation: Ensure all components are squared and then summed before taking the square root.
   • Direction Cosine Formula: Remember that $\cos \alpha$ involves the x-component divided by the magnitude of the direction vector, not just the x-component alone.
   • Scalar Multiples: Simplifying the direction vector by dividing by a common factor (as done in Step 2) does not change the direction cosines and can make calculations easier.

4. Summary

   To find the angle a line of intersection of two planes makes with an axis, we first identify the normal vectors of the planes. Their cross product gives a direction vector for the line. Finally, we use the direction cosine formula, which is the ratio of the relevant component of the direction vector to its magnitude, to find the cosine of the angle with the specified axis. Following these steps, we found the direction vector to be $(3, -3, 3)$ (or simplified to $(1, -1, 1)$), and its magnitude to be $\sqrt{3}$. The cosine of the angle $\alpha$ with the positive x-axis is $\frac{1}{\sqrt{3}}$.

The final answer is $\boxed{\text{A}}$