Find the distance from the origin (0,0,0) to the plane ABC, where A=(1,1,0); B=(0,1,3); C=(-1,3,3)?

The distance from (0,0,0) to the plane ax+by+cz+d=0 is √(d²/(a² + b² + c²))
Here a = (By-Ay)(Cz-Az) - (Bz-Az)(Cy-Ay), b = (Bz-Az)(Cx-Ax) - (Bx-Ax)(Cz-Az), c = (Bx-Ax)(Cy-Ay) - (By-Ay)(Cx-Ax), and d = aAx + bAy + cAz = aBx + bBy + cBz = aCx + bCy + cCz (choose one of A, B or C)

Here:
a = (1-1)(3-0) - (3-0)(3-1) = 0 - 6 = -6
b = (3-0)(-1-1) - (0-1)(3-0) = -6 - -3 = -9
c = (0-1)(3-1) - (1-1)(-1-1) = -2 - 0 = -2
and (choosing point A) d = a + b = -6 + -9 = -15.

So the distance is √((-15)²/((-6)² + (-9)² + (-2)²)) = 15/11 ⇦ ⇦ ⇦ 𝔸𝕟𝕤𝕨𝕖𝕣

...at the near point (a,b,c)*d/(a²+b²+c²) which is (90/121, 135/121, 30/121) - not too far away from point A, which is understandable given point A is significantly closer to the origin than either point B or C.

See: http://mathworld.wolfram.com/Point-PlaneDistance.html