JEE Main 2026 — Three Dimensional Geometry Question with Solution

Let the coordinates of $P$ on the first line be $(2\lambda ,3\lambda ,3\lambda -1)$.

Let the coordinates of $Q$ on the second line be $(-\mu -1,\mu +2,4\mu )$.

The direction ratios of the line segment $PQ$ are proportional to $1,-1,2$. Thus, we can write:

$\frac{-\mu -1-2\lambda }{1}=\frac{\mu +2-3\lambda }{-1}=\frac{4\mu -3\lambda +1}{2}=k$

This gives the system of equations:

$-\mu -2\lambda -1=k$

$\mu -3\lambda +2=-k$

$4\mu -3\lambda +1=2k$

Adding the first two equations gives:

$-5\lambda +1=0\Rightarrow \lambda =\frac{1}{5}$

Substituting $\lambda =\frac{1}{5}$ into the first and third equations:

$-\mu -\frac{7}{5}=k\Rightarrow \mu +k=-\frac{7}{5}$

$4\mu +\frac{2}{5}=2k\Rightarrow 2\mu -k=-\frac{1}{5}$

Adding these two equations gives:

$3\mu =-\frac{8}{5}\Rightarrow \mu =-\frac{8}{15}$

Substituting $\mu =-\frac{8}{15}$ into $\mu +k=-\frac{7}{5}$:

$-\frac{8}{15}+k=-\frac{21}{15}\Rightarrow k=-\frac{13}{15}$

The vector $\overrightarrow{PQ}$ is $(k,-k,2k)$. The length of the line segment $PQ$ is $\alpha =\sqrt{k^2+(-k{)}^2+(2k{)}^2}=\sqrt{6k^2}$.

Thus, ${\alpha }^2=6k^2=6{\left(-\frac{13}{15}\right)}^2=6\times \frac{169}{225}=\frac{1014}{225}$.

Therefore, $225{\alpha }^2=1014$.

Answer: $1014$