Question

Equation of the plane passing through the point $(3,-1,2)$ and parallel to the lines $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+1}{3}$ and $\frac{x-4}{3}=\frac{y}{-1}=\frac{z}{2}$ is

• A. $x+y-z=2$
• B. $x+y-z=0$
• C. $x+y-z=3$
• D. $x+y+z+3=0$

Answer 1

The correct option is B $x+y-z=0$

Equation of the plane passing through

$(3,-1,2)$ is given by,

$A(x-3)+B(y+1)+C(z-2)=0$ ---- $\left(1\right)$

where $A,B,C$ are the direction ratios of normal to the plane

Since, plane $\left(1\right)$ is parallel to given lines

therefore,

$2A+B+3C=0$ ----- $\left(2\right)$

$3A-B+2C=0$ ---- $\left(3\right)$

On addind $2$ and $3$

we get,

$A=-C$

putting

$A=-C$ in $1$, we get

$B=-C$

putting $A=-C$ and $B=-C$ in $\left(1\right)$

we get ,

$x+y-z=0$

Hence required equation of the plane is $x+y-z=0$

then,

the correct option is $B$.