Given x + 2Y + 3Z = 12, find the minimum value of x ^ 2 + 2Y ^ 2 + 3Z ^ 2

Given x + 2Y + 3Z = 12, find the minimum value of x ^ 2 + 2Y ^ 2 + 3Z ^ 2

By Cauchy inequality
(x^2+2y^2+3z^2)(1+2+3)>=(x+2y+3z)^2=144
So the minimum is 24