Three Dimensional Geometry – Straight Lines – JEE Main 24 Jan 2026 Shift 2

Solution:

Line 1: Point $A(-1, 2, 4)$, Vector $\vec{b_1} = \alpha\hat{i} – \hat{j} – \alpha\hat{k}$

Line 2: Point $B(0, 1, 1)$, Vector $\vec{b_2} = \alpha\hat{i} + 2\hat{j} + 2\alpha\hat{k}$

Vector $\vec{AB} = (0 – (-1))\hat{i} + (1 – 2)\hat{j} + (1 – 4)\hat{k} = \hat{i} – \hat{j} – 3\hat{k}$

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{vmatrix} = \hat{i}(-2\alpha + 2\alpha) – \hat{j}(2\alpha^2 + \alpha^2) + \hat{k}(2\alpha + \alpha)$

$\vec{b_1} \times \vec{b_2} = 0\hat{i} – 3\alpha^2\hat{j} + 3\alpha\hat{k}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{9\alpha^4 + 9\alpha^2} = 3|\alpha|\sqrt{\alpha^2+1}$

Shortest Distance formula: $SD = \frac{|\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$

Numerator: $|1(0) + (-1)(-3\alpha^2) + (-3)(3\alpha)| = |3\alpha^2 – 9\alpha| = 3|\alpha^2 – 3\alpha|$

Given $SD = \sqrt{2}$:
$$\sqrt{2} = \frac{3|\alpha^2 – 3\alpha|}{3|\alpha|\sqrt{\alpha^2+1}} = \frac{|\alpha||\alpha – 3|}{|\alpha|\sqrt{\alpha^2+1}}$$

$$\sqrt{2} = \frac{|\alpha – 3|}{\sqrt{\alpha^2+1}}$$

Squaring both sides:
$$2(\alpha^2 + 1) = (\alpha – 3)^2$$

$$2\alpha^2 + 2 = \alpha^2 – 6\alpha + 9$$

$$\alpha^2 + 6\alpha – 7 = 0$$

$$(\alpha + 7)(\alpha – 1) = 0 \Rightarrow \alpha = -7, 1$$

Sum of values = $-7 + 1 = -6$.

Ans. (2)