1. Given $a, b, c$ are all positive integers, and satisfy the conditions
$$
a^{2}-b^{2}-c^{2}=a b c, \quad a^{2}=2(b+c) .
$$

Find $a=$ $\qquad$ ,$b=$ $\qquad$ ,$\dot{c}=$ $\qquad$
$$
\begin{array}{l}
\text { II. 1. } 2,1,1 . \\
\because a>0, b>0, c>0, \\
\therefore a b c>0, \text { i.e., } a^{2}-b^{2}-c^{2}>0 . . \\
\text { Thus, } a>b, a>c . \\
\text { Also, } \because a^{2}=2(b+c)<4 a, \\
\therefore a<4 .
\end{array}
$$

From $a^{2}=2(b+c)$, we know that $a$ is an even number,
$$
\therefore a=2, b=c=1 \text {. }
$$