Let the shortest distance between the lines $L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$ and $L_{1}: x+1=y-1=4-z$ be $2 \sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$, then which of the following is NOT possible?

Question

Let the shortest distance between the lines

$L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}, \lambda \geq 0$ and

$L_{1}: x+1=y-1=4-z$ be $2 \sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$,

then which of the following is NOT possible?

$\alpha+2 \gamma=24$

$2 \alpha+\gamma=7$

$\alpha-2 \gamma=19$

$2 \alpha-\gamma=9$

Solution

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