Q42 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
Let the lines $\mathrm{L}_{1}: \vec{r}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}), \lambda \in \mathbb{R}$ and $\mathrm{L}_{2}: \vec{r}=(4 \hat{\mathrm{i}}+\hat{\mathrm{j}})+\mu(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$, intersect at the point R. Let P and Q be the points lying on lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, respectively, such that $|\overrightarrow{\mathrm{PR}}|=\sqrt{29}$ and $|\overrightarrow{\mathrm{PQ}}|=\sqrt{\frac{47}{3}}$. If the point P lies in the first octant, then $27(\mathrm{QR})^{2}$ is equal to
- A348
- B340
- C320
- Correct answerD360
Solution
Find R by solving $L_1 = L_2$:
From $1+2\lambda = 4+5\mu$, $2+3\lambda = 1+2\mu$, $3+4\lambda = \mu$ we get $\lambda = -1$, $\mu = -1$, so $R = (-1, -1, -1)$
Point P on $L_1$: $P = (1+2\lambda_P, 2+3\lambda_P, 3+4\lambda_P)$ with $|\overrightarrow{PR}|^2 = 29$
$4(1+\lambda_P)^2 + 9(1+\lambda_P)^2 + 16(1+\lambda_P)^2 = 29$ → $29(1+\lambda_P)^2 = 29$ → $\lambda_P = 0$ (first octant)
Thus $P = (1, 2, 3)$
Point Q on $L_2$: $Q = (4+5\mu_Q, 1+2\mu_Q, \mu_Q)$ with $|\overrightarrow{PQ}|^2 = \frac{47}{3}$
$(3+5\mu_Q)^2 + (-1+2\mu_Q)^2 + (\mu_Q-3)^2 = \frac{47}{3}$
$30\mu_Q^2 + 20\mu_Q + 19 = \frac{47}{3}$ → $9\mu_Q^2 + 6\mu_Q + 1 = 0$ → $\mu_Q = -\frac{1}{3}$
So $Q = (\frac{7}{3}, \frac{1}{3}, -\frac{1}{3})$
$|QR|^2 = \frac{100+16+4}{9} = \frac{120}{9} = \frac{40}{3}$
$27(QR)^2 = 360$
From $1+2\lambda = 4+5\mu$, $2+3\lambda = 1+2\mu$, $3+4\lambda = \mu$ we get $\lambda = -1$, $\mu = -1$, so $R = (-1, -1, -1)$
Point P on $L_1$: $P = (1+2\lambda_P, 2+3\lambda_P, 3+4\lambda_P)$ with $|\overrightarrow{PR}|^2 = 29$
$4(1+\lambda_P)^2 + 9(1+\lambda_P)^2 + 16(1+\lambda_P)^2 = 29$ → $29(1+\lambda_P)^2 = 29$ → $\lambda_P = 0$ (first octant)
Thus $P = (1, 2, 3)$
Point Q on $L_2$: $Q = (4+5\mu_Q, 1+2\mu_Q, \mu_Q)$ with $|\overrightarrow{PQ}|^2 = \frac{47}{3}$
$(3+5\mu_Q)^2 + (-1+2\mu_Q)^2 + (\mu_Q-3)^2 = \frac{47}{3}$
$30\mu_Q^2 + 20\mu_Q + 19 = \frac{47}{3}$ → $9\mu_Q^2 + 6\mu_Q + 1 = 0$ → $\mu_Q = -\frac{1}{3}$
So $Q = (\frac{7}{3}, \frac{1}{3}, -\frac{1}{3})$
$|QR|^2 = \frac{100+16+4}{9} = \frac{120}{9} = \frac{40}{3}$
$27(QR)^2 = 360$
Hence, the correct answer is option: 'D'.