JEE Main 2026 — Mathematics Vectors & 3D Geometry

The given lines are $L_1: \frac{x+1}{\alpha} = \frac{y-2}{-1} = \frac{z-4}{-\alpha}$L1​:αx+1​=−1y−2​=−αz−4​ and $L_2: \frac{x}{\alpha} = \frac{y-1}{2} = \frac{z-1}{2\alpha}$L2​:αx​=2y−1​=2αz−1​.

For $L_1$L1​, point $\vec{a}_1 = (-1, 2, 4)$a1​=(−1,2,4) and direction $\vec{b}_1 = (\alpha, -1, -\alpha)$b1​=(α,−1,−α).

For $L_2$L2​, point $\vec{a}_2 = (0, 1, 1)$a2​=(0,1,1) and direction $\vec{b}_2 = (\alpha, 2, 2\alpha)$b2​=(α,2,2α).

The shortest distance $d$d is given by $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$d=∣b1​×b2​∣∣(a2​−a1​)⋅(b1​×b2​)∣​.

$\vec{a}_2 - \vec{a}_1 = (1, -1, -3)$a2​−a1​=(1,−1,−3).

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{vmatrix} = \hat{i}(-2\alpha + 2\alpha) - \hat{j}(2\alpha^2 + \alpha^2) + \hat{k}(2\alpha + \alpha) = (0, -3\alpha^2, 3\alpha)$b1​×b2​=​i^αα​j^​−12​k^−α2α​​=i^(−2α+2α)−j^​(2α2+α2)+k^(2α+α)=(0,−3α2,3α).

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{0 + 9\alpha^4 + 9\alpha^2} = 3|\alpha|\sqrt{\alpha^2 + 1}$∣b1​×b2​∣=0+9α4+9α2​=3∣α∣α2+1​.

$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(0) + (-1)(-3\alpha^2) + (-3)(3\alpha) = 3\alpha^2 - 9\alpha$(a2​−a1​)⋅(b1​×b2​)=(1)(0)+(−1)(−3α2)+(−3)(3α)=3α2−9α.

Given $d = \sqrt{2}$d=2​, so $\frac{|3\alpha^2 - 9\alpha|}{3|\alpha|\sqrt{\alpha^2 + 1}} = \sqrt{2} \Rightarrow \frac{|\alpha - 3|}{\sqrt{\alpha^2 + 1}} = \sqrt{2}$3∣α∣α2+1​∣3α2−9α∣​=2​⇒α2+1​∣α−3∣​=2​.

Squaring both sides: $(\alpha - 3)^2 = 2(\alpha^2 + 1) \Rightarrow \alpha^2 - 6\alpha + 9 = 2\alpha^2 + 2$(α−3)2=2(α2+1)⇒α2−6α+9=2α2+2.

$\alpha^2 + 6\alpha - 7 = 0$α2+6α−7=0.

The values of $\alpha$α are the roots of this quadratic equation.

Sum of values of $\alpha = -\frac{6}{1} = -6$α=−16​=−6.