JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Three Dimensional Geometry

The midpoint of $P(a,6,9)$ and its image $(20,b,-a-9)$ is $\left(\frac{a+20}{2},\frac{b+6}{2},-\frac{a}{2}\right)$ and it lies on the given line, so for some $t$ it equals $(3+7t,2+5t,1-9t)$. The three coordinates give $a=14t-14$, $b=10t-2$ and $a=18t-2$. Equating the two expressions for $a$ gives $14t-14=18t-2$, so $t=-3$, and then $a=-56$ and $b=-32$. The vector from $P$ to its image is $(76,-38,38)$, and $7(76)+5(-38)-9(38)=0$, so it is perpendicular to the direction $(7,5,-9)$ as a mirror image requires. Hence $|a+b|=|-56-32|=88$.