2}{1} \\ & (x, y, z)=(\lambda+1,2 \lambda+3, \lambda+2) \\ & \text { Put in } L_1: x-y+3 z=6 \\ & (\lambda+1)-(2 \lambda+3)+3(\lambda+2)=6 \\ & 2 \lambda=2 \Rightarrow \lambda=1 \\ & Q=(2,5,3) \end{aligned}$$

line: $$\frac{x-2}{1}=\frac{y-5}{-1}=\frac{z-3}{3}$$

$$(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\mathrm{t}+2,5-\mathrm{t}, 3 \mathrm{t}+3)$$

Put in $$L_2: 2 x-y+z=-4$$

$$\begin{aligned} & 2(t+4)-(5-t)+(3 t+3)=-4 \\ & 6 t=-6 \Rightarrow t=-1 \\ & R=(1,6,0) \end{aligned}$$

JEE Advanced 2024 Paper 2 Online Mathematics - 3D Geometry Question 5 English Explanation

$$\begin{aligned} & \text { Perimeter }=\sqrt{6}+\sqrt{13}+\sqrt{11} \\ & \text { Centroid }=\left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}\right) \end{aligned}$$

19:T565,

$$\begin{aligned} & \text { रेखा }: \frac{x-1}{1}=\frac{y-3}{2}=\frac{z-2}{1} \\ & (x, y, z)=(\lambda+1,2 \lambda+3, \lambda+2) \\ & \text { समीकरण में रखने पर } L_1: x-y+3 z=6 \\ & (\lambda+1)-(2 \lambda+3)+3(\lambda+2)=6 \\ & 2 \lambda=2 \Rightarrow \lambda=1 \\ & Q=(2,5,3) \end{aligned}$$

रेखा: $$\frac{x-2}{1}=\frac{y-5}{-1}=\frac{z-3}{3}$$

$$(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\mathrm{t}+2,5-\mathrm{t}, 3 \mathrm{t}+3)$$

समीकरण में रखने पर $$L_2: 2 x-y+z=-4$$

$$\begin{aligned} & 2(t+4)-(5-t)+(3 t+3)=-4 \\ & 6 t=-6 \Rightarrow t=-1 \\ & R=(1,6,0) \end{aligned}$$

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