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

x^2+2y^2+3z^2)(1+2+3)>=(x+2y+3z)^2=144
So x ^ 2 + 2Y ^ 2 + 3Z ^ 2 > = 144 / 6 = 24
That is, the minimum value of x ^ 2 + 2Y ^ 2 + 3Z ^ 2 is 24

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