\begin{align*} \text{By Cauchy-Schwarz}\\ \ (\frac{1}{x}+\frac{4}{y}+\frac{9}{z})(x+y+z)\geq (1+2+3)^2\\ \text{Then}\\ \ 12\geq(x+y+z)\geq12\\ \text{Since equality occurs if and only if}\\ \ \frac{x}{\frac{1}{x}}=\frac{y}{\frac{4}{y}}=\frac{z}{\frac{9}{z}}\\ \ x=\frac{y}{2}=\frac{z}{3}\\ \text{x=1 y=2 z=3}\\ \end{align*}

One more solution is (x, y, z) = (2, 4, 6).

(x, y, z) = (1, 2, 3) is not solution.