Complicated Lagrange multipliers problem

Question

I'm having a lot of trouble solving this problem. I got $$ \frac {192}{w+12}$$ but it is wrong. Here is the problem:

Answer 1

$4\ln U=3\log x_1+\log x_2$

$x_1+x_2+x_3=24$

$12x_1-wx_3=0$

So we have linear dependence of $(\frac{3}{x_1},\frac{1}{x_2},0), (1,1,1), (12,0,-w) $ and therefore of

$(\frac{3}{x_1},\frac{1}{x_2},0)$ and $(12+w,w,0)$. Then $$\frac{w}{12+w}=\frac{x_1}{3x_2}.$$ Therefore

$$\left(\frac{w}{12}+\frac{12+w}{36}+1\right)x_3=24$$ $$x_3=\frac{216}{w+12}.$$