The equation of the plane, passing through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and…

The equation of the plane, passing through the intersection of the planes $x+y+z=1$ and $2 x+3 y-z+4=0$ and parallel to $Y$-axis is
  1. $x+4 z-1=0$
  2. $x+4 z-7=0$
  3. $x-4 z+7=0$
  4. $x-4 z+1=0$

Solution

Equation of plane passing through the intersection of given planes is $\begin{aligned} & (x+y+z-1)+\lambda(2 x+3 y-z+4)=0 \\ & \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+4 \lambda-1=0 \end{aligned}$
Since the plane is parallel to Y -axis. $\begin{aligned} \therefore \quad & 1+3 \lambda=0 \\ & \Rightarrow \lambda=\frac{-1}{3} \end{aligned}$ $\therefore \quad$ Equation of the required plane is $x+4 \mathrm{z}-7=0$

Asked in: MHT CET 2024 (16 May Shift 1)

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