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−1λλεR is equal to k633 then k is equal to ____

x+12=y−34=z+13=λ1→2λ1−1, 4λ1+3, 3λ1−1x+32=y+26=z−1λ=p→2p−3 , 6p−2 λp+1λ1=−12 p=12, λPoint −2, 1, −52Distance =23−2−101−2−52+48232+−102+−22=3633=k633k=3

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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 $\frac{x + 3}{2} = \frac{y + 2}{6} = \frac{z - 1}{λ}$ $λ ε R$ is equal to $\frac{k}{\sqrt{633}}$ then $k$ is equal to ____

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answer is 3.

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Detailed Solution

$\frac{x + 1}{2} = \frac{y - 3}{4} = \frac{z + 1}{3} = λ_{1}$

$\to (2 λ_{1} - 1, 4 λ_{1} + 3, 3 λ_{1} - 1)$

$\frac{x + 3}{2} = \frac{y + 2}{6} = \frac{z - 1}{λ} = p$

$\to (2 p - 3, 6 p - 2 λ p + 1)$

$λ_{1} = \frac{- 1}{2} p = \frac{1}{2}, λ$

Point $(- 2, 1, - \frac{5}{2})$

Distance $= \frac{|23 (- 2) - 10 (1) - 2 (\frac{- 5}{2}) + 48|}{\sqrt{{(23)}^{2} + {(- 10)}^{2} + {(- 2)}^{2}}}$ $= \frac{3}{\sqrt{633}}$

$= \frac{k}{\sqrt{633}}$ $k = 3$

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