smc

The first equation suggests the substitution $(a,b,c)=(x,2y,4z),$ from which $(x,y,z)=\left(a,\frac{b}{2},\frac{c}{4}\right).$ We rewrite the given equations in terms of $a,b,$ and $c:$ $$\begin{array}{rl}a+b+c& =12,\\ \frac{ab}{2}+\frac{bc}{2}+\frac{ac}{2}& =22,\\ \frac{abc}{8}& =6.\end{array}$$ We clear fractions in these equations: $$\begin{array}{rl}a+b+c& =12,\\ ab+ac+bc& =44,\\ abc& =48.\end{array}$$ By Vieta's Formulas, note that $a,b,$ and $c$ are the roots of the equation $$r^3-12r^2+44r-48=0,$$ which factors as $$(r-2)(r-4)(r-6)=0.$$ It follows that $\{a,b,c\}=\{2,4,6\}.$ Since the substitution $(x,y,z)=\left(a,\frac{b}{2},\frac{c}{4}\right)$ is not symmetric with respect to $x,y,$ and $z,$ we conclude that different ordered triples $(a,b,c)$ generate different ordered triples $(x,y,z),$ as shown below: \begin{array}{c|c|c||c|c|c} & & & & & \\ [-2.5ex] \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} & \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{z} \\ [0.5ex] \hline & & & & & \\ [-2ex] 2 & 4 & 6 & 2 & 2 & 3/2 \\ 2 & 6 & 4 & 2 & 3 & 1 \\ 4 & 2 & 6 & 4 & 1 & 3/2 \\ 4 & 6 & 2 & 4 & 3 & 1/2 \\ 6 & 2 & 4 & 6 & 1 & 1 \\ 6 & 4 & 2 & 6 & 2 & 1/2 \end{array} So, there are $\boxed{6}$ such ordered triples $(x,y,z).$