Let $π_{1} : 2 x - 3 y + 6 z + 8 = 0$ and $π_{2} : 3 x - 2 y - 2 z + 7 = 0$ be two planes and $A (0, \frac{29}{9}, \frac{5}{18})$ lies on both these planes. Two points $B (2, 0, - 2)$ and $C$ are such that the line of intersection of the planes is the internal angle bisector of $∠ A$ of $Δ A B C$ and $A B = A C .$ Which is/are correct

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If the co-ordinates of $C$ are $(α, β, γ)$ then $β + γ$ equals 6

If the co-ordinates of $C$ are $(α, β, γ)$ then $α + 2 γ$ equals 0

The x-intercept of the plane containing the point $C$ and perpendicular to planes $π_{1}$ and $π_{2}$ is $\frac{13}{9}$

The y-intercept of the plane containing the point $C$ and perpendicular to planes $π_{1}$ and $π_{2}$ is $\frac{13}{11}$

answer is A, B, D, C.

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Detailed Solution

$\begin{array}{l} \underline{\begin{matrix} 2 & - 3 & 6 \\ 3 & - 2 & - 2 \end{matrix}} \\ 18 22 5 \\ C o m m o n l i n e : \frac{x - 0}{18} = \frac{y - \frac{29}{9}}{22} = \frac{z - \frac{5}{18}}{5} = t \end{array}$

$\begin{array}{l} F o o t (18 t, 22 t + \frac{29}{9}, 5 t + \frac{5}{18}) \\ B D ⊥ A D \Rightarrow (18 t - 2, 22 t + \frac{29}{9}, 5 t + \frac{41}{18}) . (18, 22, 5) = 0 \\ \Rightarrow 324 t - 36 + 484 t + \frac{29}{9} \times 22 + 25 t + \frac{205}{18} = 0 \\ \Rightarrow 833 t - 36 + \frac{481}{18} = 0 \Rightarrow t = - \frac{1}{18} \\ D = (- 1, 2, 0) \Rightarrow C = (- 4, 4, 2) \\ P k_{a} : 18 x + 22 y + 5 z = 26 \end{array}$