Let $a,$ $b,$ $c$ be real numbers such that
\[a + 3b + 4c = a^2 + b^2 + c^2 = 25.\]Find the largest possible value of $a.$
By Cauchy-Schwarz,
\[(3^2 + 4^2)(b^2 + c^2) \ge (3b + 4c)^2.\]Since $a + 3b + 4c = a^2 + b^2 + c^2 = 25,$ we can write this as
\[25(25 - a^2) \ge (25 - a)^2.\]Expanding, we get
\[625 - 25a^2 \ge 625 - 50a + a^2,\]so $26a^2 - 50a \le 0.$  This factors as $2a(13a - 25) \le 0.$  This implies $a \le \frac{25}{13}.$

For $a = \frac{25}{13},$ since we have equality above, we want $\frac{b^2}{9} = \frac{c^2}{16}.$  We also want $a + 3b + 4c = 25.$  We can solve to get $b = \frac{36}{13}$ and $c = \frac{48}{13},$ so the largest possible value of $a$ is $\boxed{\frac{25}{13}}.$