The direction cosines of the line $x-y+z-5=0=x-3 y-6=0$ are:

$\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{-2}{\sqrt{14}}$

The correct answer is Option (3) → $\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, \frac{-2}{\sqrt{14}}$

1. Identify the Normal Vectors

The equations of the two planes are:

• Plane 1: $x - y + z - 5 = 0 \rightarrow$ Normal vector $\vec{n_1} = (1, -1, 1)$
• Plane 2: $x - 3y - 6 = 0 \rightarrow$ Normal vector $\vec{n_2} = (1, -3, 0)$

2. Find the Direction Ratios (D.R.s)

The direction ratios $(a, b, c)$ of the line of intersection are given by the cross product of the normal vectors: $\vec{d} = \vec{n_1} \times \vec{n_2}$.

Using the determinant method:

$\vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & -3 & 0 \end{vmatrix}$

• i-component: $(-1)(0) - (1)(-3) = 0 + 3 = 3$
• j-component: $-[(1)(0) - (1)(1)] = -(0 - 1) = 1$
• k-component: $(1)(-3) - (-1)(1) = -3 + 1 = -2$

So, the direction ratios are $(3, 1, -2)$.

3. Find the Direction Cosines (D.C.s)

To convert direction ratios $(a, b, c)$ to direction cosines $(l, m, n)$, we divide by the magnitude:

$\text{Magnitude } = \sqrt{a^2 + b^2 + c^2} = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$

The direction cosines are:

$l = \frac{3}{\sqrt{14}}, \quad m = \frac{1}{\sqrt{14}}, \quad n = \frac{-2}{\sqrt{14}}$

Final Answer:

The direction cosines are $(\frac{3}{\sqrt{14}}, \frac{1}{\sqrt{14}}, -\frac{2}{\sqrt{14}})$.