#### Answer

The solution of the system is $\left( 0,0,4 \right)$.

#### Work Step by Step

Let us consider the given equation:
$x+3y+5z=20$ ...... (I)
$y-4z=-16$ ...... (II)
$3x-2y+9z=36$ ...... (III)
Now, multiply equation (I) by 3 and subtract from equation (III):
$\begin{align}
& 3x+9y+15z=6 \\
& 3x-2y+9z=36 \\
\end{align}$
Solve and we get:
$11y+6z=24$ ...... (IV)
Multiply equation (II) by 11 and subtract from equation (IV):
$\begin{align}
& 11y+6z=24 \\
& 11y-44z=-176 \\
\end{align}$
Solve and we get:
$\begin{align}
& 50z\ =\ 200 \\
& z=4
\end{align}$
Now, putting the value of $z=4$ in equation (II):
$\begin{align}
& y-4\times 4=-16 \\
& y=0
\end{align}$
Putting the value of $y=0$ and $z=4$ in equation (I)
$\begin{align}
& x+3\left( 0 \right)+5\left( 4 \right)=20 \\
& x+20=20 \\
& x=0
\end{align}$
Thus, the ordered triple $\left( 0,0,4 \right)$ satisfies the three systems.