Write an equation of the plane with normal vector n =⟨−7,−4,−6⟩ passing through the point (−8,4,6) in scalar form answer box =20

The equation of a plane in 3D space is given by the formula:

A(x - x0) + B(y - y0) + C(z - z0) = 0

where (A, B, C) is the normal vector of the plane, and (x0, y0, z0) is a point on the plane.

In this case, the normal vector n =⟨−7,−4,−6⟩, and the point is (−8,4,6). So, A = -7, B = -4, C = -6, x0 = -8, y0 = 4, and z0 = 6.

Substituting these values into the equation of the plane, we get:

-7(x - (-8)) - 4(y - 4) - 6(z - 6) = 0

Simplify this to get the equation of the plane in scalar form:

-7x + 56 - 4y + 16 - 6z + 36 = 0

Combine like terms:

-7x - 4y - 6z + 108 = 0

So, the equation of the plane in scalar form is -7x - 4y - 6z + 108 = 0.