AB=(-2，5，0)，AC=(-2，0，3)

Let n=(x, y, z) be the normal vector of plane ABC.

Then n ab = 0, n AC = 0.

So {-2x+5y=0.

{-2x+3z=0

Solution: y = 2/5 * x, z = 2/3 * x.

Then n=(x, 2/5*x, 2/3*x},

X= 1:

n=( 1，2/5，2/3)

Do not set x=0. At this time, n = (0,0,0) is worthless.