JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Three Dimensional Geometry

Take the foot of the perpendicular from $P(a,4,2)$ to the line as $F=(2\lambda -1,3\lambda +3,1-\lambda )$. Requiring $\overrightarrow{PF}$ perpendicular to the direction $(2,3,-1)$ gives $14\lambda -4-2a=0$, so $\lambda =\frac{a+2}{7}$. Imposing $|\overrightarrow{PF}{|}^2=24$ reduces to $5a^2+6a-155=0$, whose positive root is $a=5$; then $\lambda =1$ and the foot is $F=(1,6,0)$. The image is the reflection of $P$ in this foot, $Q=2F-P=(-3,8,-2)$, so $\sum {\alpha }_i=-3+8-2=3$ and $a+{\sum }_{i=1}^3{\alpha }_i=5+3=8$.