Question

A plane parallel to $y-$axis passing through line of intersection of planes $x+y+z=1$ and $2x+3y-z-4=0$ which of the point lie on the plane

• A. $(3,2,1)$
• B. $(-3,0,1)$
• C. $(-3,1,1)$
• D. $(3,1,-1)$

Answer 1

The correct option is D

$(3,1,-1)$

Explanation for the correct option:

Step 1: Find the equation of the plane:

Use the two given planes to find the plane equation.

$$\begin{gathered} \Rightarrow \left(x+y+z-1\right)+\lambda \left(2x+3y-z-4\right)=0 \\ \Rightarrow \left(1+2\lambda \right)x+(1+3\lambda )y+(1-\lambda )z-(1+4\lambda )=0 \end{gathered}$$

Step 2: Find the value of $\lambda$.

Since it is parallel to $y$ axis

$$\begin{gathered} \Rightarrow 1+3\lambda =0 \\ \Rightarrow \lambda =\frac{-1}{3} \end{gathered}$$

So the plane equation is $x+4z+1=0$

On verifying the point by physically putting them in the equation of the plane, only $(3,1,-1)$ satisfies the equation.

Hence, option (d) is the correct answer.