$17 \cdot 181$ For all triangles with integer side lengths and a perimeter of 12, the incorrect conclusion is
(A) There must be an equilateral triangle.
(B) There must be a right triangle.
(C) There must be an isosceles right triangle.
(D) Each one must be an isosceles triangle or a right triangle.
(3rd "Five Sheep Cup" Junior High School Mathematics Competition, 1991)
[Solution] Let the three sides of the triangle be $a, b, c$, and $a \leqslant b \leqslant c$. From $a+b+c \leqslant 3 c$, i.e., $12 \leqslant 3 c$, we know $c \geqslant 4$.
Also, $a+b+c=(a+b)+c>2 c$, i.e., $12>2 c$, we know $c<6$. Therefore, $c=4$ or 5.
Thus, the side lengths of the triangle that meet the conditions are only three:
$$
(4,4,4),(3,4,5),(2,5,5) \text {. }
$$

None of them are isosceles right triangles.
Therefore, the answer is $(C)$.