Finding the value of an expression given three equations. $$x+y+2z=22$$
$$3x-2y+z=6$$
$$7x+3y-5z=1$$
above are three equations and we have to find the value of $$x^2+y^2+z^2=?$$
 A: Hint: $$\det \begin{vmatrix}1&1&2\\3&-2&1\\7&3&-5\end{vmatrix}\neq 0$$ so the system has a unique solution. Find it and substitute in $x^2+y^2+z^2$. To check you should find $(x,y,z)=(3,5,7)$. Can you solve the system of three equations with the three unknowns (by performing elementary row operations)?
A: $$
\begin{cases}
x+y+2z=22\\
3x-2y+z=6\\
7x+3y-5z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
3x-2y+z=6\\
7x+3y-5z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
3x-2(22+(-x-2z))+z=6\\
7x+3(22+(-x-2z))-5z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
-44+5x+5z=6\\
66+4x-11z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=10-z\\
66+4x-11z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=10-z\\
66+4(10-z)-11z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=10-z\\
106-15z=1
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=10-z\\
z=7
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=10-7\\
z=7
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-x-2z)\\
x=3\\
z=7
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=22+(-3-2\cdot7)\\
x=3\\
z=7
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
y=5\\
x=3\\
z=7
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
x=3\\
y=5\\
z=7
\end{cases}
$$

So:
$$x^2+y^2+z^2=3^2+5^2+7^2=9+25+49=83$$
