Points A, B, C, and T are in space such that each of TA, TB, and TC is perpendicular to the other two. If TA = TB = 12 and TC = 6, then what is the distance from T to face ABC?

Answer 1

The equation of the plane through A,B,C is
x/12 + y/12 + z/6 = 1
or
x+y+2z = 12
Now use the formula for the distance from (0,0,0) to x+y+2z-12 = 0
12/√(1^2 + 1^2 + 2^2) = 12/√6