Assume $a \geq b$ by symmetry; then $0 < c < b \leq a < 99$. Also $2a^2 \geq a^2 + b^2 > 99^2$, so that $a \geq 71$. Thus $71 \leq a \leq 98$. Write the equation as $(b + c)(b - c) = (99 - a)(99 + a)$.

For $\min(a + b + c)$ look at $a = 98$ and $a = 97$ first. If $a = 98$ then $(b + c)(b - c) = 197$, and $197$ is a prime. Hence $b + c = 197$, $a + b + c = 295$. If $a = 97$ then $(b + c)(b - c) = 2 \cdot 196 = 392 = 14 \cdot 28$, and $392$ has no representation $392 = d_1 d_2$ with $d_1, d_2$ between $14$ and $28$. The sum of integers with a fixed product is a minimum when the factors are as close as possible, therefore $b + c \geq 28$ and $a + b + c \geq 97 + 28 = 125$. The equality $a + b + c = 125$ is attained for $b + c = 28$, $b - c = 14$, i.e. $b = 21$, $c = 7$; the triple $(97, 21, 7)$ is admissible. We show that $125$ is the desired minimum.

Note that $(b + c)(b - c) = (99 - a)(99 + a)$ implies $b + c > \sqrt{99^2 - a^2}$, hence $a + b + c > a + \sqrt{99^2 - a^2}$. So a sufficient condition for $a + b + c > 125$ is $a + \sqrt{99^2 - a^2} > 125$. This is equivalent to $a^2 - 125a + 2912 < 0$. The quadratic function $f(t) = t^2 - 125t + 2912$ increases in $[62\frac{1}{2}, +\infty)$ and $f(94) < 0$, implying $a + b + c > 125$ for $a \in [71, 94]$. If $a = 95$ then $(b + c)(b - c) = 4 \cdot 194$. As $b + c > b - c$ and $b + c$, $b - c$ have the same parity, we see that $b + c \geq 194 > 125$. If $a = 96$ then $(b + c)(b - c) = 3 \cdot 195 = 15 \cdot 39$, and the factors in the last product are closest possible. Hence $b + c \geq 39$ and $a + b + c \geq 96 + 39 = 135 > 125$.

For the maximum of $a + b + c$ note first that $a + b + c$ is odd. Also $99 - a < b - c < b + c < 99 + a$ in view of $0 < c < b \leq a < 99$ and $(b + c)(b - c) = (99 - a)(99 + a)$. Moreover the four numbers $b + c$, $b - c$, $99 - a$, $99 + a$ have the same parity, so $b - c = (99 - a) + 2k$ with $k \geq 1$ an integer. We prove that the maximum is attained when $k = 1$, i.e. $b - c = 101 - a$.

First restrict attention to admissible triples $(a, b, c)$ that satisfy the last additional condition. Set $x = b - c = 101 - a$ for clarity. Then $a = 101 - x$, and one can express $a + b + c$ in terms of $x$:
$$a + b + c = a + \frac{(99 - a)(99 + a)}{b - c} = 303 - 2\left(x + \frac{200}{x}\right).$$
Since $a + b + c$ is odd, $x + \frac{200}{x}$ is an integer. So $x$ and $\frac{200}{x}$ are divisors of $200$ with product $200$. We have to minimize their sum in order that $a + b + c$ be a maximum. A factorization $d_1 d_2 = 200$ is needed with $d_1$ and $d_2$ closest possible, which leads to $\{x, \frac{200}{x}\} = \{10, 20\}$. Therefore $a + b + c \leq 303 - 2(10 + 20) = 243$. The value $243$ is attained for the triple $(91, 81, 71)$ which is admissible.

If $b - c \ne 101 - a$ then $b - c \geq (99 - a) + 4 = 103 - a > 0$, hence
$$a + b + c = a + \frac{(99 - a)(99 + a)}{b - c} \leq a + \frac{(99 - a)(99 + a)}{103 - a}.$$
To complete the proof it suffices to show that $a + \frac{(99 - a)(99 + a)}{103 - a} < 243$ whenever $a < 99$. Use the last inequality to reach the equivalent form $a^2 - 73a + 7614 > 0$. The latter holds for all real $a$ since the trinomial $a^2 - 73a + 7614$ has a negative discriminant.