3D Geometry – Lines – JEE Main 21 Jan 2026 Shift 2

Solution:

Equation of line $L_1$ passing through $(2, 6, 7)$ parallel to $-3\hat{i}+2\hat{j}+4\hat{k}$:
$$\frac{x-2}{-3}=\frac{y-6}{2}=\frac{z-7}{4} = \lambda_1$$
Any point $C$ on $L_1$ is $(-3\lambda_1+2, 2\lambda_1+6, 4\lambda_1+7)$.

Equation of line $L_2$ passing through $(4, 3, 5)$ parallel to $2\hat{i}+\hat{j}+3\hat{k}$:
$$\frac{x-4}{2}=\frac{y-3}{1}=\frac{z-5}{3} = \lambda_2$$
Any point $D$ on $L_2$ is $(2\lambda_2+4, \lambda_2+3, 3\lambda_2+5)$.

The direction ratios of the line segment $CD$ are:
$$((2\lambda_2+4)-(-3\lambda_1+2), (\lambda_2+3)-(2\lambda_1+6), (3\lambda_2+5)-(4\lambda_1+7))$$
$$(2\lambda_2+3\lambda_1+2, \lambda_2-2\lambda_1-3, 3\lambda_2-4\lambda_1-2)$$

Since $L_3$ (line $CD$) is parallel to $-3\hat{i}+5\hat{j}+16\hat{k}$, their direction ratios are proportional:
$$\frac{2\lambda_2+3\lambda_1+2}{-3} = \frac{\lambda_2-2\lambda_1-3}{5} = \frac{3\lambda_2-4\lambda_1-2}{16}$$

From the first two terms:
$$5(2\lambda_2+3\lambda_1+2) = -3(\lambda_2-2\lambda_1-3)$$
$$10\lambda_2+15\lambda_1+10 = -3\lambda_2+6\lambda_1+9$$
$$13\lambda_2+9\lambda_1+1 = 0 \quad \dots (1)$$

From the last two terms:
$$16(\lambda_2-2\lambda_1-3) = 5(3\lambda_2-4\lambda_1-2)$$
$$16\lambda_2-32\lambda_1-48 = 15\lambda_2-20\lambda_1-10$$
$$\lambda_2-12\lambda_1-38 = 0 \Rightarrow \lambda_2 = 12\lambda_1+38 \quad \dots (2)$$

Substitute (2) into (1):
$$13(12\lambda_1+38) + 9\lambda_1 + 1 = 0$$
$$156\lambda_1 + 494 + 9\lambda_1 + 1 = 0$$
$$165\lambda_1 = -495 \Rightarrow \lambda_1 = -3$$

Substitute $\lambda_1 = -3$ back into (2):
$$\lambda_2 = 12(-3) + 38 = 2$$

Now find the coordinates of $C$ and $D$:
$$C(-3(-3)+2, 2(-3)+6, 4(-3)+7) \Rightarrow C(11, 0, -5)$$
$$D(2(2)+4, 2+3, 3(2)+5) \Rightarrow D(8, 5, 11)$$

Calculate $|\overline{CD}|^2$:
$$|\overline{CD}|^2 = (11-8)^2 + (0-5)^2 + (-5-11)^2$$
$$|\overline{CD}|^2 = 3^2 + (-5)^2 + (-16)^2$$
$$|\overline{CD}|^2 = 9 + 25 + 256 = 290$$

Ans. (2)