JEE Main 2026 — Three Dimensional Geometry Question with Solution

Equating the position vectors of the two lines for intersection:

$(1+a\lambda )\hat{i}+(1-\lambda )\hat{j}-\hat{k}=(4+2\mu )\hat{i}+0\hat{j}+(-1+a\mu )\hat{k}$

Comparing the coefficients of $\hat{i},\hat{j},\hat{k}$:

$1-\lambda =0\Rightarrow \lambda =1$

$-1=-1+a\mu \Rightarrow a\mu =0$

Since $a\ne 0$, we get $\mu =0$.

Substituting $\lambda =1$ and $\mu =0$ in the $\hat{i}$ component equation:

$1+a(1)=4+2(0)\Rightarrow a=3$

The position vector of the point of intersection is obtained by substituting $\mu =0$ in the second line:

$\vec{r}=4\hat{i}-\hat{k}$

The coordinates of the point of intersection are $(4,0,-1)$.

The square of the distance from the origin is $4^2+0^2+(-1{)}^2=16+1=17$.

Answer: $17$