The thing about an equation like 2x-3y+z-6=0 is that if you look carefully, there's a dot product hiding in plain sight. The equation can also be written

2(x-0)-3(y-0)+1(z-6)=0

and also

〈2,-3,1〉•〈x-0,y-0,z-6〉= 0,

which means that if you take any point (x,y,z) in this plane, subtract the point (0,0,6), then the resulting vector is orthogonal to the vector〈2,-3,1〉. The vector 〈2,-3,1〉 called a *normal* of the plane. Conveniently, to find the minimum distance between a point and the plane, we need only subtract (0,0,6), dot with the plane normal, take the absolute value, and divide by the magnitude of the plane normal.

(-2,0,3) - (0,0,6) = 〈-2,0,-3〉

〈-2,0,-3〉 • 〈2,-3,1〉 = -7

||〈2,-3,1〉|| = ✓(14)

The distance is 7/✓(14) units.