JEE MAIN - Mathematics (2023 - 30th January Evening Shift - No. 19)
Let a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$. If the distance of $L$ from the point $(5,3,8)$ is $\alpha$, then $3 \alpha^2$ is equal to :
Answer
158
Explanation
$$L:{{x - 2} \over 1} = {{y - 3} \over { - 1}} = {{z - 1} \over { - 1}} = \lambda $$
Any point on L can be taken as
$$B(\lambda + 2, - \lambda + 3, - \lambda + 1)$$
Let $$A(5,3,8)$$
So, $$AB\,.\,(\widehat i - \widehat j - \widehat k) = 0$$
$$[(\lambda - 3)\widehat i - \lambda \widehat j - (\lambda + 7)\widehat k]\,.\,[\widehat i - \widehat j - \widehat k] = 0$$
$$\lambda - 3 + \lambda + \lambda + 7 = 0$$
$$\therefore$$ $$\lambda = {{ - 4} \over 3}$$
$$\overrightarrow {AB} = {{13} \over 3}\widehat i + {4 \over 3}\widehat i - {{17} \over 3}\widehat k$$
$$|\overrightarrow {AB} | = \sqrt {{{169} \over 9} + {{16} \over 9} + {{289} \over 9}} $$
$$ = {{\sqrt {474} } \over 3} = \alpha $$
$$3{\alpha ^2} = {{474} \over 9} \times 3 = 158$$
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