jmc

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 $\boxed{\frac{25}{13}}.$