QJEEM23Mac218: JEE Main 2023 Mathematics - Class 12 JEEM23Mac218 Solution: Let a line pass through the point and be parallel to the line . …

$L:{{x - 2} \over 1} = {{y - 3} \over { - 1}} = {{z - 1} \over { - 1}} = \lambda$

Any point on L can be taken as

$B(\lambda + 2, - \lambda + 3, - \lambda + 1)$

Let $A(5,3,8)$

So, $AB\,.\,(\widehat i - \widehat j - \widehat k) = 0$

$[(\lambda - 3)\widehat i - \lambda \widehat j - (\lambda + 7)\widehat k]\,.\,[\widehat i - \widehat j - \widehat k] = 0$

$\lambda - 3 + \lambda + \lambda + 7 = 0$

$\therefore$ $\lambda = {{ - 4} \over 3}$

$\overrightarrow {AB} = {{13} \over 3}\widehat i + {4 \over 3}\widehat i - {{17} \over 3}\widehat k$

$|\overrightarrow {AB} | = \sqrt {{{169} \over 9} + {{16} \over 9} + {{289} \over 9}}$

$= {{\sqrt {474} } \over 3} = \alpha$

$3{\alpha ^2} = {{474} \over 9} \times 3 = 158$