1. Determine the smallest values that the expression $V=(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$ can take, given that the real numbers $a$, $b$, $c$ satisfy the pair of conditions

$$
\begin{aligned}
a+3 b+c & =6 \\
-a+b-c & =2
\end{aligned}
$$
SOLUTION. By adding both equations, we find that $b=2$. Substituting $b$ into one of them yields $c=-a$. Therefore, $V=(a-2)^{2}+(2+a)^{2}+(-2 a)^{2}$. After squaring and adding, we find that $V=6 a^{2}+8 \geqq 8$. Equality occurs precisely when $a=0, b=2$ and $c=0$.

The sought minimum value of the expression $V$ is thus equal to 8.

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