JEE Main 2026 — Jan 24, Shift 1 — Mathematics solution

JEE Main2026Jan 24, Shift 1Mathematics

Let the lines $L_{1} : r = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$ , $λ \in R$ and $L_{2} : r = (4 \hat{i} + \hat{j}) + μ (5 \hat{i} + 2 \hat{j} + \hat{k})$ , $μ \in R$ , intersect at the point $R$ . Let $P$ and $Q$ be the points lying on lines $L_{1}$ and $L_{2}$ , respectively, such that $∣ P R ∣ = 29$ and $∣ P Q ∣ = \frac{47}{3}$ . If the point $P$ lies in the first octant, then $27 (QR)^{2}$ is equal to

A
$340$
B
$360$
C
$320$
D
$348$

Solution

Find $R$ : Equate components of $L_{1}$ and $L_{2}$ .

$1 + 2 λ = 4 + 5 μ, 2 + 3 λ = 1 + 2 μ, 3 + 4 λ = μ$ .

From the third, $μ = 4 λ + 3$ ; substituting in the second gives $λ = - 1$ , $μ = - 1$ .

$\Rightarrow R = (- 1, - 1, - 1)$ .

Find $P$ : Let $P = (1 + 2 t, 2 + 3 t, 3 + 4 t)$ on $L_{1}$ .

$∣ P R ∣^{2} = (2 t + 2)^{2} + (3 t + 3)^{2} + (4 t + 4)^{2} = (t + 1)^{2} (4 + 9 + 16) = 29 (t + 1)^{2} = 29$ .

$\Rightarrow (t + 1)^{2} = 1 \Rightarrow t = 0$ or $t = - 2$ . The first-octant point is $t = 0$ , giving $P = (1, 2, 3)$ .

Find $Q$ : Let $Q = (4 + 5 s, 1 + 2 s, s)$ on $L_{2}$ .

$∣ P Q ∣^{2} = (5 s + 3)^{2} + (2 s - 1)^{2} + (s - 3)^{2} = 30 s^{2} + 20 s + 19 = \frac{47}{3}$ .

$\Rightarrow 90 s^{2} + 60 s + 10 = 0 \Rightarrow 9 s^{2} + 6 s + 1 = 0 \Rightarrow (3 s + 1)^{2} = 0 \Rightarrow s = - \frac{1}{3}$ .

$\Rightarrow Q = (\frac{7}{3}, \frac{1}{3}, - \frac{1}{3})$ .

Compute $27 (QR)^{2}$ :

$(QR)^{2} = (\frac{7}{3} + 1)^{2} + (\frac{1}{3} + 1)^{2} + (- \frac{1}{3} + 1)^{2} = \frac{100 + 16 + 4}{9} = \frac{120}{9}$ .

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

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