When divided by $7$ a perfect square can give the remainder $0$, $1$, $2$ or $4$. For $a^2 + b^2 + c^2$ to be divisible by $7$, the numbers $a^2$, $b^2$ and $c^2$ must either all give the remainder $0$, or they must give three different remainders, $1$, $2$ and $4$.

In the first case the numbers $a$, $b$, $c$ are divisible by $7$, so $a^4 + b^4 + c^4$ is divisible by $7$ as well.

In the second case we may assume that $a^2 = 7k + 1$, $b^2 = 7m + 2$ and $c^2 = 7n + 4$ for some integers $k$, $m$ and $n$. In this case
$$a^4 + b^4 + c^4 = (7k+1)^2 + (7m+2)^2 + (7n+4)^2 = 49(k^2+m^2+n^2)+14(k+m+n)+21,$$
which is again divisible by $7$.