Question

Let the plane containing the line of intersection of the planes $P 1: x+(\lambda+4) y+z=1$ and $P 2: 2 x+$ $y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$ Then the distance of the point $(2 \lambda, \lambda,-\lambda)$ from the plane $P 2$ is

• A. $5 \sqrt{6}$
• B. $2 \sqrt{6}$
• C. $3 \sqrt{6}$
• D. $4 \sqrt{6}$

Answer 1

The Correct Option is C

The correct answer is (C) : $3 \sqrt{6}$
Equation of plane passing through point of intersection of $P1$ and $P2$
$P=P1+kP2$
$(x+(\lambda +4)y+z-1)+k(2x+y+z-2)=0$
Passing through $(0,1,0)$ and $(1,0,1)$
$(\lambda +4-1)+k(1-2)=0$
$(\lambda +3)-k=\mathrm{0.....}$(1)
Also passing $(1,0,1)$
$(1+1-1)+k(2+1-2)=0$
$1+k=0$
$k=-1$
put in (1)
$\lambda +3+1=0$
$\lambda =-4$
Then point $(2\lambda ,\lambda ,-\lambda )$
$d=\left|\frac{-16-4,-4,4)}{\sqrt{6}}\right|$
$d=\frac{18}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}=3\sqrt{6}$

Answer 2

Step 1: Equation of the plane passing through the line of intersection of \( P_1 \) and \( P_2 \) 

The equation of the required plane can be written as:

\[ P = P_1 + kP_2, \] where \( P_1 \) is: \[ x + (\lambda + 4)y + z - 1 = 0 \] and \( P_2 \) is: \[ 2x + y + z - 2 = 0. \] Thus, the equation of the plane is: \[ x + (\lambda + 4)y + z - 1 + k(2x + y + z - 2) = 0. \] Simplifying: \[ (1 + 2k)x + (\lambda + 4 + k)y + (1 + k)z - (1 + 2k) = 0. \] 

Step 2: Passing through the points \( (0, 1, 0) \) and \( (1, 0, 1) \)

Substitute \( (0, 1, 0) \) into the plane equation:

\[ 0 + (\lambda + 4 + k)(1) + 0 - (1 + 2k) = 0. \] Simplify: \[ \lambda + 4 + k - 1 - 2k = 0 \quad \Rightarrow \quad \lambda + 3 - k = 0 \quad \text{(Equation 1)}. \]

Substitute \( (1, 0, 1) \) into the plane equation:

\[ (1 + 2k)(1) + (\lambda + 4 + k)(0) + (1 + k)(1) - (1 + 2k) = 0. \] Simplify: \[ 1 + 2k + 1 + k - 1 - 2k = 0 \quad \Rightarrow \quad 1 + k = 0. \] Solve: \[ k = -1. \] 

Step 3: Solve for \( \lambda \) Substitute \( k = -1 \) into Equation 1: \[ \lambda + 3 - (-1) = 0 \quad \Rightarrow \quad \lambda + 3 + 1 = 0. \] Therefore: \[ \lambda = -4. \] 

Step 4: Find the point \( (2\lambda, \lambda, -\lambda) \) Using \( \lambda = -4 \): \[ (2\lambda, \lambda, -\lambda) = (2(-4), -4, -(-4)) = (-8, -4, 4). \] 

Step 5: Distance of \( (-8, -4, 4) \) from the plane \( P_2 \) The distance of a point \( (x_1, y_1, z_1) \) from a plane \( ax + by + cz + d = 0 \) is given by the formula: \[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}. \] For \( P_2 : 2x + y + z - 2 = 0 \), substitute \( a = 2 \), \( b = 1 \), \( c = 1 \), and \( d = -2 \): \[ d = \frac{|2(-8) + 1(-4) + 1(4) - 2|}{\sqrt{2^2 + 1^2 + 1^2}}. \] Simplify: \[ d = \frac{|-16 - 4 + 4 - 2|}{\sqrt{4 + 1 + 1}} = \frac{|-18|}{\sqrt{6}} = \frac{18}{\sqrt{6}}. \] Rationalize: \[ d = \frac{18\sqrt{6}}{6} = 3\sqrt{6}. \]