JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Three Dimensional Geometry

Line $L_1$ through $(1,2,3)$ parallel to the z-axis has direction $\overrightarrow{d_1}=(0,0,1)$, and $L_2$ through $(\lambda ,5,6)$ parallel to the y-axis has direction $\overrightarrow{d_2}=(0,1,0)$. Their cross product is $\overrightarrow{d_1}\times \overrightarrow{d_2}=(-1,0,0)$, and the vector joining the base points is $(\lambda -1,3,3)$. The shortest distance is $\frac{|-(\lambda -1)|}{1}=|\lambda -1|$, and setting this equal to $3$ gives $\lambda =4$ or $\lambda =-2$. Since ${\lambda }_2<{\lambda }_1$, ${\lambda }_1=4$ and ${\lambda }_2=-2$, so the point is $P(4,-2,7)$. As $L_1$ is the vertical line $x=1,y=2$, the foot of the perpendicular from $P$ is $(1,2,7)$ and the distance is $\sqrt{(4-1{)}^2+(-2-2{)}^2}=\sqrt{9+16}=5$. Hence the square of the distance of $({\lambda }_1,{\lambda }_2,7)$ from $L_1$ is $25$.