JEE Main 2026 — 24 January, Morning Shift — Question 3

For POI $\begin{aligned} & 2 \lambda+1=5 \mu+4 ; 3 \lambda+2=2 \mu+1 ; 4 \lambda+3=\mu & \end{aligned}$

$\Rightarrow \lambda=\mu=-1$

$\mathrm{R}(-1,-1,-1) \quad \mathrm{P}(2 \lambda+1,3 \lambda+2,4 \lambda+3)$

$\mathrm{PR}^{2}=29 \Rightarrow(2 \lambda+2)^{2}+(3 \lambda+3)^{2}+(4 \lambda+4)^{2}=29$

$\Rightarrow \lambda=0 \text { or } \lambda=-2 \text { (Reject) }$

$\Rightarrow \mathrm{P}(1,2,3)$

$\mathrm{Q}(5 \mu+4,2 \mu+1, \mu)$

$|\mathrm{PQ}|=\sqrt{\frac{47}{3}} \Rightarrow \mathrm{PQ}^{2}=\frac{47}{3}$

$\Rightarrow(5 \mu+3)^{2}+(2 \mu-1)^{2}+(\mu-3)^{2}=\frac{47}{3}$

$\Rightarrow \mu=-\frac{1}{3}$

$\mathrm{Q}=\left(\frac{7}{3}, \frac{1}{3},-\frac{1}{3}\right)$

$(\mathrm{QR})^{2}=\left(\frac{7}{3}+1\right)^{2}+\left(\frac{1}{3}+1\right)^{2}+\left(-\frac{1}{3}+1\right)^{2}$

$=\frac{100+16+4}{9}=\frac{120}{9}$

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