JEE MAIN - Mathematics (2025 - 7th April Evening Shift - No. 14)

Consider the lines L1: x - 1 = y - 2 = z and L2: x - 2 = y = z - 1. Let the feet of the perpendiculars from the point P(5, 1, -3) on the lines L1 and L2 be Q and R respectively. If the area of the triangle PQR is A, then 4A2 is equal to :
151
147
139
143

Explanation

$$\begin{aligned} & \mathrm{L}_1: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-0}{2} \\ & \text { Let } \mathrm{Q}(\lambda+1, \lambda+2, \lambda) \\ & \overrightarrow{\mathrm{PQ}}=(\lambda-4, \lambda-1, \lambda+3) \\ & \overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{~m}}=0 \end{aligned}$$

JEE Main 2025 (Online) 7th April Evening Shift Mathematics - 3D Geometry Question 15 English Explanation

$$\begin{aligned} &\begin{aligned} & \Rightarrow \lambda-4+\lambda+1, \lambda+3=0 \\ & \Rightarrow 3 \lambda=0 \\ & \Rightarrow \lambda=0 \\ & \Rightarrow \mathrm{Q}(1,2,0) \\ & \mathrm{L}_2: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-1}{2} \end{aligned}\\ &\text { Let } \mathrm{R}(\mu+2, \mu, \mu+1) \overrightarrow{\mathrm{PR}}=(\mu-3, \mu-1, \mu+4)\\ &\overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{n}}=0\\ &\mu-3+\mu-1+\mu+4=0\\ &\neq \mu=0\\ &\mathrm{R}(2,0,1) \end{aligned}$$

$$\begin{aligned} & \text { Area of } \triangle \mathrm{PQR}(\mathrm{~A})=\frac{1}{2}|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}| \\ & \mathrm{A}=\frac{1}{2}|(-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}})| \\ & \mathrm{A}=\frac{1}{2}|7(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})| \\ & \left|\begin{array}{lrr} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -4 & 1 & 3 \\ -3 & -1 & 4 \end{array}\right| \\ & =7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & 4 \mathrm{~A}^2=49 \times 3=147 \end{aligned}$$

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