Find the cartesian equation of the plane through the point(2,5,3) with normal 3i+2j-7k.

The normal is: n̂ = 3î + 2ĵ – 7k̂.

Let P1 and P be two position vectors of points on the plane, where

P1 is the position vector of the given point on the plane, (2, 5, 3), and

P is the position vector of a general point on the plane, (x, y, z).

Let R be a vector contained within the plane such that R̂ = P̂ – P̂1.

Since n̂ is the normal vector to the plane, then n̂•R̂ = 0.

i.e. n̂•(P̂ – P̂1) = 0

or, n̂•P̂ = n̂•P̂1

Expanding,

(3î + 2ĵ – 7k̂)•(xî + yĵ + zk̂) = (3 2 -7)•(2 5 3)

3x + 2y – 7z = 3*2 + 2*5 – 7*3 = 6 + 10 – 21 = -5

__Therefore, equation of the plane is: 3x + 2y – 7z + 5 = 0__