The answer is (A). The condition that $a + b + c$ has the same number of digits as $a^{2} + b^{2} + c^{2}$ implies that
$$\frac{a^{2} + b^{2} + c^{2}}{a + b + c} < 10$$
Therefore we must have that
$$a^{2} - 10a + b^{2} - 10b + c^{2} - 10c < 0$$
which is equivalent to stating that
$$(a - 5)^{2} + (b - 5)^{2} + (c - 5)^{2} < 75.$$
In particular $(c - 5)^{2} < 75$, from which we obtain $c < 14$. Let us now observe that if $9 < c < 14$ then $a^{2} + b^{2} + c^{2}$ has at least 3 digits, while $a + b + c \leq 3c$ has 2. Hence $c$ is at most 9, and it is easy to check that the triple $(1, 2, 9)$ satisfies the required conditions.