$32 \cdot 13$ The sum of the squares of three consecutive natural numbers is 2 more than 8 times their sum. Then the sum of the squares of these three natural numbers is
(A) 77.
(B) 149.
(C) 194.
(D) 245.
(3rd "Five Sheep Cup" Junior High School Mathematics Competition, 1991)
[Solution] Let the three consecutive natural numbers be $x-1, x, x+1$. According to the problem, we have
$$
(x-1)^{2}+x^{2}+(x+1)^{2}=8[(x-1)+x+(x+1)]+2 \text {, }
$$

which simplifies to $3 x^{2}+2=24 x+2$, or $x^{2}-8 x=0$.
Solving this, we get $x=8$ or 0 (discard 0).
Thus, the sum of the squares of the three numbers is $7^{2}+8^{2}+9^{2}=194$.
Therefore, the answer is $(\mathrm{C})$.