Question

Equation of the plane through $(1,-1,2)$ and perpendicular to the planes $x+2y+2z=5$ and $3x+3y+2z=8$ is

• A. $2x-4y+3z-12=0$
• B. $2x+4y+3z-4=0$
• C. $2x+4y-3z+8=0$
• D. $2x-4y-3z=0$

Answer 1

The correct option is B $2x+4y+3z-4=0$

$x+2y+2z=5$

$3x+3y+2z=8(1,-1,2)$

Let $a(x-1)+b(y+1)+c(2-2)=0\dots \dots \left(1\right)$

be perpendicular to the planes

$x+2y+2z=5$ & $3x+3y+2z=8$

$\therefore a+2b+2c=5$ &

$3a+3b+2c=8$

Solving the above equation by cross

Multiplication , we ,get;

$\frac{a}{4-6}=\frac{b}{2-6}=\frac{c}{6-3}$

$\frac{a}{-2}=\frac{b}{-4}=\frac{c}{-3}$

$\Rightarrow a=-2,b=-4,c=-3$

Substituting the above in equation$\dots \dots \left(1\right)$

$-2(x-1)-4(y+1)-3(z-2)=0$

$-2x+2-4y-4-3z+6=0$

$-2x-4y-3z+2+2=0$

$-2x-4y-3z+4=0$

$2x+4y+3z=4$

Is the equation of the plane solution

$\therefore$ option$\left(b\right)$ is correct.