JEE Main 2025 Apr 4 Shift 1, Mathematics Q6: Lines in Space

Set up two points and the direction vectors. Take a point on each line by choosing the parameter zero:

$A=(3,\alpha ,3)$ on line 1 with direction $\vec{p}=(3,-1,1)$,

$B=(-3,-7,\beta )$ on line 2 with direction $\vec{q}=(-3,2,4)$.

Compute the common perpendicular direction $\vec{p}\times \vec{q}$.

$\vec{p}\times \vec{q}=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 3-11\\ -324\end{array}\right|=(-4-2)\hat{i}-(12-(-3))\hat{j}+(6-3)\hat{k}=-6\hat{i}-15\hat{j}+3\hat{k}$.

Its magnitude: $|\vec{p}\times \vec{q}|=\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}$.

Apply the shortest-distance formula.

$d=\frac{|\overrightarrow{AB}\cdot (\vec{p}\times \vec{q})|}{|\vec{p}\times \vec{q}|}$, with $\overrightarrow{AB}=B-A=(-6,-7-\alpha ,\beta -3)$.

Numerator (before taking the absolute value):

$\overrightarrow{AB}\cdot (\vec{p}\times \vec{q})=(-6)(-6)+(-7-\alpha )(-15)+(\beta -3)(3)$

$=36+15(\alpha +7)+3(\beta -3)=15\alpha +3\beta +132$.

Use the given shortest distance. We are told $d=3\sqrt{30}$, and we already found $|\vec{p}\times \vec{q}|=3\sqrt{30}$, so:

$d=\frac{|15\alpha +3\beta +132|}{3\sqrt{30}}=3\sqrt{30}\Rightarrow |15\alpha +3\beta +132|=270$.

Resolving the absolute value. The absolute value gives two cases: $15\alpha +3\beta +132=270$ or $15\alpha +3\beta +132=-270$.

Case 1: $15\alpha +3\beta =138$, dividing by 3 gives $5\alpha +\beta =46$.

Case 2: $15\alpha +3\beta =-402$, dividing by 3 gives $5\alpha +\beta =-134$.

The question asks for the positive value of $5\alpha +\beta$, so we take Case 1.

$\therefore 5\alpha +\beta =\boxed{46}$.