Question -8

the plane is parallel to another plane $2x-2y+z = 0$ which mean that it also have similar form but gotta be added with a constant $c$ . we can say that the equation of our plane is $2x-2y+z+c = 0$ .

to get the value of c, we substitute a point that's passed by the plane : substitue $(1,-2,3)$ to $2x-2y+z+c = 0$ , we get $c = -9$ .

now let's count the distance between a point $(-1,2,0)$ to the plane $2x-2y+z-9 = 0$ using the formulation:

$d = \lvert \frac{a*x_0 + b*y_0 + c*z_0+d}{a^2+b^2+c^2} \rvert$ where $(x_0,y_0,z_0)$ is the coordinate of the point.

by doing it,we get: $\boxed{d=5}$

plane passes through a point and parallel to another plane eqn is a(x-x ^)+b(y-y^)+c(z-z ^) solve and compare with ax+by+cz+d=o u will get a,b,c values then find distance ans is 5