If the shortest between the lines x+6 2 = y-6 3 = z-6 4 and x- 3 … - Vidyadip

MCQ

If the shortest between the lines $\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$ and $\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{z+2 \sqrt{6}}{5}$ is $6$, then the square of sum of all possible values of $\lambda$ is

A
$380$
B
$3885$
C
$386$
✓
$384$

✓

Answer

Correct option: D.

$384$

d
Shortest distance between the lines

$\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$

$\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}$ is $6$

Vector along line of shortest distance

$=\left|\begin{array}{lll} i & j & k \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|, \Rightarrow-\hat{ i }+2 \hat{ j }- k$ (its magnitude is $\sqrt{6}$ )

$\begin{array}{l}\text { Now } \frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\sqrt{6}+\lambda & \sqrt{6} & -3 \sqrt{6} \\2 & 3 & 4 \\3 & 4 & 5\end{array}\right|=\pm 6 \\\Rightarrow \lambda=-2 \sqrt{6}, 10 \sqrt{6}\end{array}$

So, square of sum of these values is $384$.

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