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 within 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,
$$
A=\{1,2,6\},\{1,3,5\},\{2,3,4\} \text {; }
$$

If $a+b+c=16$, then $7 \leqslant c \leqslant 12$, we can have
$$
\begin{aligned}
A= & \{1,3,12\},\{1,4,11\},\{2,3,11\}, \\
& \{1,5,10\},\{2,4,10\},\{1,6,9\}, \\
& \{2,5,9\},\{3,4,9\},\{1,7,8\}, \\
& \{2,6,8\},\{3,5,8\},\{3,6,7\}, \\
& \{4,5,7\} ;
\end{aligned}
$$

If $a+b+c=25$, then $10 \leqslant c \leqslant 12$, in this 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 $A$.