Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the equation of the plane be $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$.

Since the plane passes through $(-1,2,-2)$, the equation becomes:

$a(x+1)+b(y-2)+c(z+2)=0$

The plane also passes through $(-1,3,2)$, so this point must satisfy the equation:

$a(-1+1)+b(3-2)+c(2+2)=0$

$b+4c=0\Rightarrow b=-4c$

The plane is perpendicular to the $yz$-plane. The normal to the $yz$-plane is the $x$-axis, which has direction ratios $(1,0,0)$. Since the required plane is perpendicular to the $yz$-plane, their normals are perpendicular:

$a(1)+b(0)+c(0)=0\Rightarrow a=0$

Substituting $a=0$ and $b=-4c$ into the equation of the plane, we get:

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

Dividing by $-c$ (assuming $c\ne 0$):

$4(y-2)-(z+2)=0$

$4y-8-z-2=0$

$4y-z=10$

Answer: $4y-z=10$