Question

If the distance between the plane αx-2y+z=k and the plane containing the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) is \(\sqrt{6}\), then|k| is :

• A. 36
• B. 12
• C. 6
• D. 2\(\sqrt3\)

Answer 1

The Correct Option is C

Step 1: Find the direction vectors of the given lines 

Let the normal vectors of the given planes be:

\[ \vec{n_1} = \langle 2, 3, 4 \rangle,\quad \vec{n_2} = \langle 3, 4, 5 \rangle \]

Step 2: Find the direction vector of the line of intersection (cross product of normals)

\[ \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = (-1)\vec{i} + 2\vec{j} - \vec{k} \]

Step 3: Equation of the plane containing the given lines and passing through point (1, 2, 3)

Using vector \(-\vec{i} + 2\vec{j} - \vec{k}\) as the normal vector and point (1, 2, 3):

\[ -1(x - 1) + 2(y - 2) - 1(z - 3) = 0 \] \[ -x + 1 + 2y - 4 - z + 3 = 0 \Rightarrow x - 2y + z = 0 \]

Step 4: Find the perpendicular distance from the origin to the plane

The distance from point \( (a, 0, 0) \) where \( a = 1 \) to the plane \( x - 2y + z = 0 \) is:

\[ \text{Distance} = \frac{|1(1) + (-2)(0) + 1(0) + d|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|1 + d|}{\sqrt{6}} \]

Given that the distance is \( \sqrt{6} \), solve: \[ \frac{|1 + d|}{\sqrt{6}} = \sqrt{6} \Rightarrow |1 + d| = 6 \Rightarrow d = 5 \text{ or } d = -7 \Rightarrow |d| = 6 \]

Final Answer: Option (C): 6