Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let $O(0,0,0)$ be the origin and $P(-1,-1,2)$ be the foot of the perpendicular to the plane.

The normal vector to the plane is $\overrightarrow{OP}$, which has direction ratios $(-1-0,-1-0,2-0)=(-1,-1,2)$.

The plane passes through the point $P(-1,-1,2)$.

The equation of a plane passing through $(x_1,y_1,z_1)$ with normal direction ratios $a,b,c$ is given by $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$.

Substituting the values, we get:

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

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

$-x-1-y-1+2z-4=0$

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

Multiplying the entire equation by $-1$, we obtain:

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

Answer: $x+y-2z+6=0$