Step 1: Understanding the Concept:
The line of intersection of two planes is perpendicular to the normal vectors of both planes.
Therefore, the direction vector of the line is the cross product of the normal vectors of the two given planes.
Step 2: Key Formula or Approach:
Let the planes be $P_1$ and $P_2$ with normal vectors $\vec{n}_1$ and $\vec{n}_2$.
The direction vector of the line is $\vec{d} = \vec{n}_1 \times \vec{n}_2$.
The direction cosines $(l, m, n)$ are given by $\frac{\vec{d}}{|\vec{d}|}$.
Step 3: Detailed Explanation:
The equations of the planes are:
Plane 1: $x - y + 2z = 5 \Rightarrow$ normal vector $\vec{n}_1 = \hat{i} - \hat{j} + 2\hat{k}$.
Plane 2: $3x + y + z = 6 \Rightarrow$ normal vector $\vec{n}_2 = 3\hat{i} + \hat{j} + \hat{k}$.
The direction vector $\vec{d}$ of the line is:
\[ \vec{d} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -1 & 2
3 & 1 & 1 \end{vmatrix} \]
\[ \vec{d} = \hat{i}(-1 \cdot 1 - 2 \cdot 1) - \hat{j}(1 \cdot 1 - 2 \cdot 3) + \hat{k}(1 \cdot 1 - (-1) \cdot 3) \]
\[ \vec{d} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 + 3) \]
\[ \vec{d} = -3\hat{i} + 5\hat{j} + 4\hat{k} \]
The magnitude of the direction vector is:
\[ |\vec{d}| = \sqrt{(-3)^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2} \]
The direction cosines are:
\[ \left( \frac{-3}{5\sqrt{2}}, \frac{5}{5\sqrt{2}}, \frac{4}{5\sqrt{2}} \right) \]
Or its negative, but this exactly matches option (A).
Step 4: Final Answer:
The direction cosines are $\frac{-3}{5\sqrt{2}}, \frac{5}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}$.