If the distance between the plane, 23x−10y−2z+48=0 and the plane c – InterviewSolution

1.	If the distance between the plane, 23x−10y−2z+48=0 and the plane containing the lines x+12=y−34=z+13 and x+32=y+26=z−1λ,(λ ∈ R) is equal to k√633, then k is equal to
Answer» If the distance between the plane, $23 x - 10 y - 2 z + 48 = 0$ and the plane containing the lines $\frac{x + 1}{2} = \frac{y - 3}{4} = \frac{z + 1}{3}$ and $\frac{x + 3}{2} = \frac{y + 2}{6} = \frac{z - 1}{λ}, (λ \in R)$ is equal to $\frac{k}{\sqrt{633}},$ then $k$ is equal to

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