We show that the only solution is $a=1$, $b=4$ and $c=3$.
Let $s = b + c \ge 7$. Substituting $a = 5-b$ and $c = s-b$ the first condition gives
$$(5-b)^2 + b^2 + (s-b)^2 = 26,$$
thus
$$3b^2 - 2(s+5)b + s^2 - 1 = 0.$$
The equation has a real solution iff the discriminant $4(s+5)^2 - 12(s^2-1) \ge 0$. This yields $s^2 - 5s - 14 \le 0$, or $(s+2)(s-7) \le 0$. Since $s \ge 7$, there must be $s=7$. If we substitute to the previous equation we get
$$3b^2 - 24b + 48 = 0;$$
with the only solution $b=4$. Then $a=1$ and $c=3$.