Q37 / 75 · JEE Main 2026 (21 January Shift 2) · Mathematics
Let the line $\mathrm{L}_{1}$ be parallel to the vector $-3 \hat{i}+2 \hat{j}+4 \hat{k}$ and pass through the point (2,6,7), and the line $\mathrm{L}_{2}$ be parallel to the vector $2 \hat{i}+\hat{j}+3 \hat{k}$ and pass through the point $(4,3,5)$. If the line $\mathrm{L}_{3}$ is parallel to the vector $-3 \hat{i}+5 \hat{j}+16 \hat{k}$ and intersects the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ at the points C and D, respectively, then $|\overrightarrow{C D}|^{2}$ is equal to :
- A89
- B312
- C171
- Correct answerD290
Solution
$L_1 : \dfrac{x-2}{-3} = \dfrac{y-6}{2} = \dfrac{z-7}{4}$
Point C on $L_1$: $(-3\lambda_1 + 2, 2\lambda_1 + 6, 4\lambda_1 + 7)$
$L_2 : \dfrac{x-4}{2} = \dfrac{y-3}{1} = \dfrac{z-5}{3}$
Point D on $L_2$: $(2\lambda_2 + 4, \lambda_2 + 3, 3\lambda_2 + 5)$
Dr's of line $L_3$ :
$L_3 : \dfrac{2\lambda_2 + 3\lambda_1 + 2}{-3} = \dfrac{\lambda_2 - 2\lambda_1 - 3}{5} = \dfrac{3\lambda_2 - 4\lambda_1 - 2}{16}$
$\lambda_1 = -3, \lambda_2 = 2$
$C(11, 0, -5)$
$D(8, 5, 11)$
$|\overrightarrow{CD}|^2 = 3^2 + 5^2 + 16^2 = 290$
Point C on $L_1$: $(-3\lambda_1 + 2, 2\lambda_1 + 6, 4\lambda_1 + 7)$
$L_2 : \dfrac{x-4}{2} = \dfrac{y-3}{1} = \dfrac{z-5}{3}$
Point D on $L_2$: $(2\lambda_2 + 4, \lambda_2 + 3, 3\lambda_2 + 5)$
Dr's of line $L_3$ :
$L_3 : \dfrac{2\lambda_2 + 3\lambda_1 + 2}{-3} = \dfrac{\lambda_2 - 2\lambda_1 - 3}{5} = \dfrac{3\lambda_2 - 4\lambda_1 - 2}{16}$
$\lambda_1 = -3, \lambda_2 = 2$
$C(11, 0, -5)$
$D(8, 5, 11)$
$|\overrightarrow{CD}|^2 = 3^2 + 5^2 + 16^2 = 290$
Hence, the correct answer is option: 'D'.