How many distinct ordered triples $(x,y,z)$ satisfy the following equations?
\begin{align*} x + 2y + 4z &= 12 \\ xy + 4yz + 2xz &= 22 \\ xyz &= 6 \end{align*}
$\textbf{(A) }\text{none}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6$
The first equation suggests the substitution $(a,b,c)=(x,2y,4z),$ from which $(x,y,z)=\left(a,\frac b2,\frac c4\right).$
We rewrite the given equations in terms of $a,b,$ and $c:$
\begin{align*} a + b + c &= 12, \\ \frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\ \frac{abc}{8} &= 6. \end{align*}
We clear fractions in these equations:
\begin{align*} a + b + c &= 12, \\ ab + ac + bc &= 44, \\ abc &= 48. \end{align*}
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 $(x,y,z)=\left(a,\frac b2,\frac c4\right)$ 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 $\boxed{\textbf{(E) }6}$ such ordered triples $(x,y,z).$
~MRENTHUSIASM (credit given to AoPS)