JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Three Dimensional Geometry

Let the required line through $C(1,1,1)$ meet the first given line at $A(1+2\lambda ,-1+3\lambda ,1+4\lambda )$ and the second at $B(3+\mu ,4+2\mu ,\mu )$. Because $A$, $C$ and $B$ lie on one line, the vectors $\overrightarrow{CA}=(2\lambda ,3\lambda -2,4\lambda )$ and $\overrightarrow{CB}=(\mu +2,2\mu +3,\mu -1)$ are parallel. Equating the first and third component ratios, $\frac{\mu +2}{2\lambda }=\frac{\mu -1}{4\lambda }$, gives $2(\mu +2)=\mu -1$, so $\mu =-5$ and $\overrightarrow{CB}=(-3,-7,-6)$. The line has direction $(3,7,6)$, so its equation is $\frac{x-1}{3}=\frac{y-1}{7}=\frac{z-1}{6}$. The point $(7,15,13)$ makes each ratio equal to $2$, so it lies on this line.