Find the distance between the point (7,2,4) and the plane determined by the points A(2,5,-3), B(-2,-3,5) and C(5,3,-3).

The general equation of a plane passing through A(2,5,-3) is a(x-2)+b(y-5)+c(z+3)=0 — (1)
It will pass through B(-2,-3,5) and C(5,3,-3) if
a(-2-2)+b(-3-5)+c(5+3)=0 ⇒ -4a-8b+8c=0 ⇒ a+2b-2c=0 — (2)
and
a(5-2)+b(3-5)+c(-3+3)=0 ⇒ 3a-2b=0 — (3)
Solving (2) and (3) by cross multiplying method, we get
a/4 = b/6 = c/4 ⇒ a/2 = b/3 = c/2 = λ
So, a=2λ, b=3λ, c=2λ
Put these values of a, b and c in (1), we get
2λ(x-2)+3λ(y-5)+2λ(z+3)=0 ⇒ 2(x-2)+3(y-5)+2(z+3)=0 ⇒ 2x-4+3y-15+2z+6=0 ⇒ 2x+3y+2z-13=0
We know that the distance of a point (x1,y1,z1) from plane ax+by+cz+d=0 is given as
d=|ax1+by1+cz1+d|/√a²+b²+c²
=|2(7)+3(2)+2(4)-13|/√4+9+4 =|14+6+8-13|/√17 = 15/√17 = 15√17/17