Let the plane containing the line of intersection of the planes $P 1 : x + (λ + 4) y + z = 1$ 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 $(2 λ, λ, - λ)$ from the plane P2 is

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$2 \sqrt{6}$

$4 \sqrt{6}$

$3 \sqrt{6}$

$5 \sqrt{6}$

answer is C.

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Detailed Solution

$\begin{array}{l} P_{1} : x + (λ + 4) y + z - 1 = 0 \\ P_{2} : 2 x + y + z - 2 = 0 \end{array}$
Equation of the plane containing the line of intersection of the plane P1 and P2 is
of the from $P_{1} + {tP}_{2} = 0$
$\Rightarrow [x + (λ + 4) y + z - 1] + t [2 x + y + z - 2] = 0$
Eq. (1) pass through (0, 1, 0) & (1, 0, 1)
$\begin{aligned} i.e., [(λ + 4) + t] - 1 - 2 t = 0 \\ \Rightarrow λ - t + 3 = 0 \dots \dots \dots . (2) \end{aligned}$
Also $(1 + 2 t) + (1 + t) - 1 - 2 t = 0 \Rightarrow t = - 1 \dots . (3)$
$Eq. (2) \& (3) λ + 1 + 3 = 0 \Rightarrow λ = - 4$
Now $(2 λ, λ, - λ) = (- 8, - 4, 4)$
Then distance from (-8, -4, 4) to P2 is
$\begin{array}{l} = \frac{|{ax}_{1} + {by}_{1} + {cz}_{1} + d|}{\sqrt{a^{2} + b^{2} + c^{2}}} \\ = \frac{| 2 (- 8) + (- 4) + (4) - 2 |}{\sqrt{4 + 1 + 1}} = \frac{| - 16 - 4 + 4 - 2 |}{\sqrt{6}} = \frac{18}{\sqrt{6}} = 3 \sqrt{6} \end{array}$