1. Given
$$
\begin{array}{l}
12 a^{2}+7 b^{2}+5 c^{2} \\
\leqslant 12 a|b|-4 b|c|-16 c-16 .
\end{array}
$$

Then $a=$ $\qquad$ ,$b=$ $\qquad$ ,$c=$ $\qquad$ .
1. $a=\frac{1}{2}, b=-1, c=-2$.

From the problem, we have
$$
\begin{array}{l}
3\left(4 a^{2}-4 a|b|+b^{2}\right)+\left(4 b^{2}+4 b|c|+c^{2}\right)+ \\
4\left(c^{2}+4 c+4\right) \leqslant 0,
\end{array}
$$

which simplifies to $3(2 a-|b|)^{2}+(2 b+|c|)^{2}+4(c+2)^{2} \leqslant 0$.
Thus, $\left\{\begin{array}{l}2 a-|b|=0, \\ 2 b+|c|=0, \\ c+2=0 .\end{array}\right.$
Solving these equations, we get $a=\frac{1}{2}, b=-1, c=-2$.