JEE Main 2026 Apr 6 Shift 1, Mathematics Q15: Lines in Space

Let $Q$ be the foot of perpendicular from $P$ to $L$. Then $Q$ is the midpoint of $P$ and its image $P^{\prime }$:

$Q=\left(\frac{1+a/3}{2},\frac{6+0}{2},\frac{a+(a+c)}{2}\right)=\left(\frac{3+a}{6},3,\frac{2a+c}{2}\right)$.

Matching $y$-coordinate of $Q$ with the line ($y=1+2t$) and $x$-coordinate ($x=t$): from $y=3$, $t=1$, so $x=1$, giving $\frac{3+a}{6}=1\Rightarrow a=3$.

So $Q=(1,3,(6+c)/2)$. The direction of $L$ is $(1,2,b)$ and $\overrightarrow{PQ}=(0,-3,c/2)$ must be perpendicular to it: $-6+bc/2=0\Rightarrow bc=12$.

Also $Q$ lies on $L$: from $z=a-1+bt=2+b$ matching $(6+c)/2$: $b=(c+2)/2$, i.e. $2b-c=2$. Solving $bc=12$, $2b-c=2$ gives $b=3$, $c=4$ (taking $b>0$). So $Q=(1,3,5)$.

A general point on $L$ is $(t,2t+1,3t+2)$. Setting $SQ=2\sqrt{14}$:

$(t-1{)}^2+(2t-2{)}^2+(3t-3{)}^2=56\Rightarrow 14(t-1{)}^2=56\Rightarrow t=3$ or $-1$.

Since $\alpha >0$, take $t=3$: $S=(3,7,11)$, and $\alpha +\beta +\gamma =21$.