Let the lines \(L_{1}: \vec{r}=\hat{i}+2 \hat{j}+3 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})\), \(\lambda \in \mathbb{R}\) and \(L_{2}: \vec{r}=(4 \hat{i}+\hat{j})+\mu(5 \hat{i}+2 \hat{j}+\hat{k}), \mu \in \mathbb{R}\), 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

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Three Dimensional Geometry

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

∣ PR ∣ = 29

and

∣ PQ ∣ = \frac{47}{3}

. If the point P lies in the first octant, then

27 (QR)^{2}

is equal to

Answer: (B)

Equating the two lines gives

λ = μ = - 1

, so

R (- 1, - 1, - 1)

. For

P (1 + 2 λ, 2 + 3 λ, 3 + 4 λ)

L_{1}

P R = (λ + 1) (2, 3, 4)

and

∣ P R ∣^{2} = 29 (λ + 1)^{2} = 29

; the first-octant choice is

λ = 0

, so

P (1, 2, 3)

. For

Q (4 + 5 μ, 1 + 2 μ, μ)

L_{2}

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

, which gives

9 μ^{2} + 6 μ + 1 = 0

, so

μ = - \frac{1}{3}

and

Q (\frac{7}{3}, \frac{1}{3}, - \frac{1}{3})

. Then

QR = (- \frac{10}{3}, - \frac{4}{3}, - \frac{2}{3})

and

(QR)^{2} = \frac{120}{9}

, so

27 (QR)^{2} = 360

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option