JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Three Dimensional Geometry

Both $L_1$ and $L_2$ pass through $P=(1,2,a)$. A point $(1+3\lambda ,2+4\lambda ,a+b\lambda )$ of $L_1$ lies on $L:\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ when $3\lambda =\frac{2+4\lambda }{2}$, giving $\lambda =1$ and the foot $(4,6,4)$ with $a+b=4$. A point $(1+\mu ,2+4\mu ,a+c\mu )$ of $L_2$ lies on $L$ when $\mu =\frac{2+4\mu }{2}$, giving $\mu =-1$ and the foot $(0,-2,0)$ with $a=c$. Equal distances from $P$ to these feet give $9+16+(a-4{)}^2=1+16+a^2$, so $8a=24$ and $a=3$. Then $c=3$, $b=4-a=1$, and $a+b+c=7$.