Answer
The values of x, y and z are as follows:
$ x=\frac{8}{a}$ ; $ y=-\frac{3}{b}$ ; $ z=-\frac{5}{c}$.
Work Step by Step
Mark the given system of equations as shown below:
$ ax-by-2cz=21$ (I)
$ ax+by+cz=0$ (II)
$2ax-by+cz=14$ (III)
By eliminating y from equations (I) and (II) to get the equation in variables x and z, we get
$\begin{align}
& \text{ }ax-by-2cz=21 \\
& \underline{\text{ }ax+by+\text{ }cz=\text{ }0} \\
& \text{2}ax\text{ }-\text{ }cz=\text{ 21} \\
\end{align}$ (IV)
By multiplying equation (III) by -1 and eliminating y from equations (I) and (III), we get
$\begin{align}
& \text{ }ax-by-2cz=21 \\
& \underline{-2ax+by-cz=-14} \\
& \text{ }-ax\text{ }-3cz=\text{ }7 \\
\end{align}$ (V)
By multiplying equation (V) by 2 and adding equations (IV) and (V), we get the value of z,
$\begin{align}
& \text{ 2}ax-\text{ }cz=\text{21} \\
& \underline{-ax-6cz=14} \\
& \text{ }-7cz=35 \\
\end{align}$ (VI)
By dividing both sides by -7c, we get,
\
$\begin{align}
& \frac{-7cz}{-7c}=\frac{35}{-7c} \\
& \text{ }z=-\frac{5}{c} \\
\end{align}$
Substitute the value of z in equation (IV) to get the value of x,
$\begin{align}
& \text{2}ax-c\left( -\frac{5}{c} \right)=\text{21 } \\
& 2ax+5=21
\end{align}$
Now, subtract 5 from both sides to get:
$\begin{align}
& 2ax+5-5=21-5 \\
& 2ax=16
\end{align}$
Now, dividing both sides by 2a, we get,
$\begin{align}
& \frac{2ax}{2a}=\frac{16}{2a} \\
& \text{ }x=\frac{8}{a} \\
\end{align}$
Substitute the values of x and z in equation (I) and we get the value of y,
$\begin{align}
& a\left( \frac{8}{a} \right)-by-2c\left( -\frac{5}{c} \right)=21 \\
& 8-by+10=21 \\
& 18-by=21
\end{align}$
Now, subtract $18$ from both sides to get:
$\begin{align}
& 18-by-18=21-18 \\
& -by=3
\end{align}$
By dividing both sides by -b we get,
$\begin{align}
& \frac{by}{b}=-\frac{3}{b} \\
& \text{ }y=-\frac{3}{b} \\
\end{align}$
Hence, the values are $ x=\frac{8}{a}$, $ y=-\frac{3}{b}$, and $ z=-\frac{5}{c}$.