Let $\theta_1=\angle APB$. By the Cosine Law,
Let $\theta_2=\angle APQ$. Note that $\cos\theta_2=-\cos\theta_1$, since $\theta_1+\theta_2=180^\circ$. So by the Cosine Law,
Add up. We get
Note the pretty cancellation!
We played a certain game at $P$. Play the same game at $R$. We get
Add $(1)$ and $(2)$, and use the Pythagorean Theorem. We get
$400+2AQ^2=2AP^2+2AR^2+100$, and dividing by $2$,
Play a similar game at $Q$, but using the double triangles $AQB$ and $AQC$. We get
so $AQ^2=100$. (This calculation is more standard: it is how we obtain a formula for the length of a median, given the sides of a triangle.)
Thus $AP^2+AR^2=250$, and therefore $AP^2+AQ^2+AR^2=350$.