JEE Main 2026 — Three Dimensional Geometry Question with Solution

Let the given point be $P(-2,-8,6)$.

The distance is measured along the line $\frac{x+5}{1}=\frac{y+5}{-1}=\frac{z}{2}$, which has direction ratios $(1,-1,2)$.

The equation of the line passing through $P$ and parallel to this line is:
$\frac{x+2}{1}=\frac{y+8}{-1}=\frac{z-6}{2}=\lambda$

Any point on this line can be written as $Q(\lambda -2,-\lambda -8,2\lambda +6)$.

Since $Q$ lies on the line $\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}$, we substitute the coordinates of $Q$ into its equation:
$\frac{\lambda -2-1}{1}=\frac{-\lambda -8-1}{2}=\frac{2\lambda +6}{-1}$

$\frac{\lambda -3}{1}=\frac{-\lambda -9}{2}=\frac{2\lambda +6}{-1}$

Taking the first two parts:
$2(\lambda -3)=-\lambda -9$
$2\lambda -6=-\lambda -9$
$3\lambda =-3\Rightarrow \lambda =-1$

Substituting $\lambda =-1$ into the coordinates of $Q$, we get:
$Q(-1-2,-(-1)-8,2(-1)+6)\equiv Q(-3,-7,4)$

The square of the distance between $P(-2,-8,6)$ and $Q(-3,-7,4)$ is:
$P{Q}^2=(-3-(-2){)}^2+(-7-(-8){)}^2+(4-6{)}^2$
$P{Q}^2=(-1{)}^2+(1{)}^2+(-2{)}^2$
$P{Q}^2=1+1+4=6$

Answer: $6$