The equation of plane through the point $(2,-1,-3):$ and parallel to lines…

The equation of plane through the point $(2,-1,-3):$ and parallel to lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is
  1. $8 x+14 y+13 z-37=0$
  2. $8 x-14 y-13 z-34=0$
  3. $8 x-14 y-13 z+37=0$
  4. $8 x+14 y+13 z+37=0$

Solution

The equation of plane passing through $(2,-1,-3)$ is $\mathrm{a}(x-2)+\mathrm{b}(y+1)+\mathrm{c}(\mathrm{z}+3)=0$
Also, as the plane is parallel to the given two lines, $\begin{aligned} \therefore \quad & 3 a+2 b-4 c=0 \text { and } 2 a-3 b+2 c=0 \\ & \Rightarrow a=-8, b=-14, c=-13 \end{aligned}$ $\therefore \quad$ The equation of the required plane is $\begin{aligned} & -8(x-2)-14(y+1)-13(z+3)=0 \\ & \Rightarrow 8 x+14 y+13 z+37=0 \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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