JEE Main 2026 Apr 4 Shift 2, Mathematics Q14: Lines in Space

Image conditions: the midpoint $M$ of the two points lies on the line, and the segment between them is perpendicular to the line's direction vector $\vec{d}=(3,2,1)$.

Midpoint $M$:

$M_x=\frac{\alpha +2\alpha +1}{2}=\frac{3\alpha +1}{2}$

$M_y=\frac{2\alpha +{\alpha }^2-3\alpha }{2}=\frac{{\alpha }^2-\alpha }{2}$

$M_z=\frac{1+(\alpha -1)/2}{2}=\frac{\alpha +1}{4}$

Substitute into the line:

$\frac{M_x-2}{3}=\frac{M_y-1}{2}=\frac{M_z}{1}$.

From the first equality: $\frac{(3\alpha +1)/2-2}{3}=\frac{({\alpha }^2-\alpha )/2-1}{2}$. Cross-multiplying and simplifying yields a cubic with $\alpha =3$ as a root.

Check perpendicularity for $\alpha =3$: segment direction is $(2\alpha +1-\alpha ,{\alpha }^2-3\alpha -2\alpha ,(\alpha -1)/2-1)=(\alpha +1,{\alpha }^2-5\alpha ,(\alpha -3)/2)=(4,-6,0)$ at $\alpha =3$. Dot with $(3,2,1)=12-12+0=0$. ✓

The other roots from $M$-on-line don't satisfy the perpendicularity condition, so only $\alpha =3$ works.