Answer
$(\frac{x+2y-z}{9},\frac{-x-2y+4z}{9},\frac{2x+y-2z}{9})$
Work Step by Step
We know that $a(0,6,3)+b(3,0,3)+c(6,-3,0)=(x,y,z)$
Thus $3b+6c=x\\
6a-3c=y\\
3a+3b=z$
Thus $3b+6c+2(6a-3c)-(3a+3b)=x+2y-z \rightarrow 9a=x+2y-z \rightarrow a=\frac{x+2y-z}{9}\\$
$3\frac{x+2y-z}{9}+3b=z \rightarrow 3b=\frac{-x-2y+4z}{3} \rightarrow b=\frac{-x-2y+4z}{9}$
$6\frac{x+2y-z}{9}-3c=y \rightarrow 3c=\frac{2x+y-2z}{3} \rightarrow c=\frac{2x+y-2z}{9}$
Thus the vector is $(\frac{x+2y-z}{9},\frac{-x-2y+4z}{9},\frac{2x+y-2z}{9})$.
