1. Let the set $M=\{1,2, \cdots, 12\}$, and the three-element set $A=$ $\{a, b, c\}$ satisfies $A \subset M$, and $a+b+c$ is a perfect square. Then the number of sets $A$ is $\qquad$ .
-、1. 26
From $6 \leqslant a+b+c \leqslant 33$, we know that the square numbers in this range are $9, 16, 25$. Let's assume $a<b<c$.

If $a+b+c=9$, then the possible values for $c$ are $6, 5, 4$, in which case,
$$
\begin{array}{l} 
A=\{1,2,6\},\{1,3,5\},\{2,3,4\} ; \\
\text { If } a+b+c=16, \text { then } 7 \leqslant c \leqslant 12, \text { we can have } \\
A=\{1,3,12\},\{1,4,11\},\{2,3,11\}, \\
\{1,5,10\},\{2,4,10\},\{1,6,9\}, \\
\{2,5,9\},\{3,4,9\},\{1,7,8\}, \\
\{2,6,8\},\{3,5,8\},\{3,6,7\}, \\
\{4,5,7\} ;
\end{array}
$$

If $a+b+c=25$, then $10 \leqslant c \leqslant 12$, in which case,
$$
\begin{aligned}
A= & \{2,11,12\},\{3,10,12\},\{4,9,12\}, \\
& \{4,10,11\},\{5,8,12\},\{5,9,11\}, \\
& \{6,7,12\},\{6,8,11\},\{6,9,10\}, \\
& \{7,8,10\} .
\end{aligned}
$$

In total, we have 26 sets $A$.