It is easy to see that there is no triplet of numbers $x, y, z$ that satisfies the following system of equations. Determine values of $x, y, z$ such that if we form the square of the difference between the two sides of each equation, the sum of the four squares is as small as possible:

$$
\begin{gathered}
x+y+z=10 \\
x+y-z=7 \\
x-y+z=6 \\
-x+y+z=5
\end{gathered}
$$
There are indeed no such triples of numbers $(x, y, z)$ that satisfy the system of equations, since the sum of the last three equations is

$$
x+y+z=18
$$

and this is in contradiction with the first equation.

Forming and then appropriately transforming the desired sum $S$:

$$
\begin{gathered}
S=(x+y+z-10)^{2}+(x+y-z-7)^{2}+(x-y+z-6)^{2}+(y-x+z-5)^{2}= \\
=2(x+y)^{2}+2(x-y)^{2}-34(x+y)-2(x-y)+(z-10)^{2}+(z+7)^{2}+(z-6)^{2}+(z-5)^{2}= \\
=4 x^{2}+4 y^{2}+4 z^{2}-36 x-32 y-28 z+210= \\
=4\left\{(x-4.5)^{2}+(y-4)^{2}+(z-3.5)^{2}+4\right\}
\end{gathered}
$$

This takes its minimum value when the terms containing the variables in the curly braces are 0, i.e., if

$$
x=4.5, \quad y=4, \quad z=3.5
$$

and in this case, the values of the left-hand sides of the given equations are $12.5, 4, 3$, their "excess" relative to the right-hand side is $+2, -2, -2, -2$, and $S=16$.