On which of the following lines lies the point of intersection of the line,…
On which of the following lines lies the point of intersection of the line, $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ ?
- $\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}$
- $\frac{x-4}{1}=\frac{y-5}{1}=\frac{z-5}{-1}$
- $\frac{x-2}{2}=\frac{y-3}{2}=\frac{z+3}{3}$
- $\frac{x+3}{3}=\frac{4-y}{3}=\frac{z+1}{-2}$
Solution
Let $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}=\lambda$
$\Rightarrow x=2 \lambda+4, y=2 \lambda+5, z=\lambda+3$
Given line lies on plane
$x+y+z=2$
$\begin{aligned} \therefore \quad & (2 \lambda+4,2 \lambda+5, \lambda+3) \text { lies on } \\ & x+y+\mathrm{z}=2 \\ & \Rightarrow(2 \lambda+4+2 \lambda+5+\lambda+3)=2 \\ & \Rightarrow 5 \lambda+12=2 \\ & \Rightarrow \lambda=-2\end{aligned}$
$\therefore \quad(0,1,1)$ lies on required plane.
$(0,1,1)$ satisfies option (A)
i.e., $\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}$
Asked in: MHT CET 2024 (03 May Shift 2)
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