To determine the direction ratios of the line of intersection of the given planes, we need to find the normal vectors of both planes first and then take their cross product.
The equations of the planes are:
Plane 1: \(x - y + z - 5 = 0\)
Plane 2: \(x - 3y - 6 = 0\)
The normal vector \(\mathbf{N}_1\) to the first plane (\(x - y + z - 5 = 0\)) is given by the coefficients of \(x\), \(y\), and \(z\), i.e., \((1, -1, 1)\).
The normal vector \(\mathbf{N}_2\) to the second plane (\(x - 3y - 6 = 0\)) is given by the coefficients of \(x\), \(y\), and \(z\), i.e., \((1, -3, 0)\).
The direction ratios of the line of intersection of two planes are given by the cross product of their normals:
\[ \mathbf{D} = \mathbf{N}_1 \times \mathbf{N}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & -3 & 0 \\ \end{vmatrix} \]
Calculating this determinant, we find:
\[ \mathbf{D} = \mathbf{i} \cdot (-1 \cdot 0 - 1 \cdot (-3)) - \mathbf{j} \cdot (1 \cdot 0 - 1 \cdot 1) + \mathbf{k} \cdot (1 \cdot (-3) - (-1) \cdot 1) \]
\[ \mathbf{D} = \mathbf{i} \cdot (0 + 3) - \mathbf{j} \cdot (0 - 1) + \mathbf{k} \cdot (-3 + 1) \]
\[ \mathbf{D} = 3\mathbf{i} + \mathbf{j} - 2\mathbf{k} \]
Thus, the direction ratios are \(3, 1, -2\).
Based on the above calculations, the correct answer should be:
Therefore, the correct direction ratios of the line of intersection of the given planes, initially thought to be \(1, -1, 1\), are indeed \(3, 1, -2\).