Three Dimensional Geometry – Lines – JEE Main 24 Jan 2026 Shift 1

Solution:

First, we find the intersection point R of lines $L_1$ and $L_2$.
General point on $L_1$: $\vec{r}_1 = (1+2\lambda)\hat{i} + (2+3\lambda)\hat{j} + (3+4\lambda)\hat{k}$
General point on $L_2$: $\vec{r}_2 = (4+5\mu)\hat{i} + (1+2\mu)\hat{j} + \mu\hat{k}$

Equating coordinates to find intersection:
$1+2\lambda = 4+5\mu \Rightarrow 2\lambda – 5\mu = 3$ …(i)
$2+3\lambda = 1+2\mu \Rightarrow 3\lambda – 2\mu = -1$ …(ii)
$3+4\lambda = \mu$ …(iii)

Substituting (iii) into (i):
$2\lambda – 5(3+4\lambda) = 3 \Rightarrow 2\lambda – 15 – 20\lambda = 3 \Rightarrow -18\lambda = 18 \Rightarrow \lambda = -1$.
From (iii), $\mu = 3 + 4(-1) = -1$.
Check with (ii): $3(-1) – 2(-1) = -3 + 2 = -1$ (Satisfied).

So, point R corresponds to $\lambda = -1$.
$R = (1-2, 2-3, 3-4) = (-1, -1, -1)$.

Point P lies on $L_1$. Let P correspond to parameter $\lambda_P$.
Distance $PR = \sqrt{29}$.
Since P and R are on the same line $L_1$, the distance is $|\lambda_P – \lambda_R| \cdot |\text{direction vector}|$.
Vector along $L_1$ is $\vec{v} = 2\hat{i}+3\hat{j}+4\hat{k}$. Magnitude $|\vec{v}| = \sqrt{4+9+16} = \sqrt{29}$.
$\vec{PR} = (\lambda_P – (-1))\vec{v} = (\lambda_P+1)\vec{v}$.
$PR^2 = (\lambda_P+1)^2 |\vec{v}|^2 \Rightarrow 29 = (\lambda_P+1)^2 (29)$.
$(\lambda_P+1)^2 = 1 \Rightarrow \lambda_P+1 = \pm 1$.
$\lambda_P = 0$ or $\lambda_P = -2$.

If $\lambda_P = 0$, $P = (1, 2, 3)$ (First Octant – Valid).
If $\lambda_P = -2$, $P = (-3, -4, -5)$ (Not First Octant).
So, $P(1, 2, 3)$.

Point Q lies on $L_2$. Let Q correspond to parameter $\mu$.
$Q = (4+5\mu, 1+2\mu, \mu)$.
Given $PQ^2 = \frac{47}{3}$.
$$(4+5\mu – 1)^2 + (1+2\mu – 2)^2 + (\mu – 3)^2 = \frac{47}{3}$$
$$(5\mu+3)^2 + (2\mu-1)^2 + (\mu-3)^2 = \frac{47}{3}$$
$$25\mu^2 + 30\mu + 9 + 4\mu^2 – 4\mu + 1 + \mu^2 – 6\mu + 9 = \frac{47}{3}$$
$$30\mu^2 + 20\mu + 19 = \frac{47}{3}$$

Multiply by 3:
$$90\mu^2 + 60\mu + 57 = 47$$
$$90\mu^2 + 60\mu + 10 = 0$$
$$9\mu^2 + 6\mu + 1 = 0 \Rightarrow (3\mu+1)^2 = 0 \Rightarrow \mu = -\frac{1}{3}$$

Coordinates of Q:
$Q = (4 – \frac{5}{3}, 1 – \frac{2}{3}, -\frac{1}{3}) = (\frac{7}{3}, \frac{1}{3}, -\frac{1}{3})$.

We need $27(QR)^2$.
$R = (-1, -1, -1)$.
$$QR^2 = \left(\frac{7}{3} + 1\right)^2 + \left(\frac{1}{3} + 1\right)^2 + \left(-\frac{1}{3} + 1\right)^2$$
$$QR^2 = \left(\frac{10}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + \left(\frac{2}{3}\right)^2$$
$$QR^2 = \frac{100}{9} + \frac{16}{9} + \frac{4}{9} = \frac{120}{9}$$

$$27(QR)^2 = 27 \times \frac{120}{9} = 3 \times 120 = 360$$

Ans. (2)