1. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
3 x-3 \geqslant 6 x+a, \\
x \geqslant 1
\end{array}\right.
$$

the solution is $1 \leqslant x \leqslant 3$. Then $a=$
$=, 1 .-12$.
From the given, we have $1 \leqslant x \leqslant \frac{1}{3}(-a-3)$. Then $\frac{1}{3}(-a-3)=3 \Rightarrow a=-12$.