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First note that $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2*(ab + bc + ca),$ and thus

$a^{2} + b^{2} + c^{2} = 6^{2} - 2*9 = 18.$

Now by Cauchy's Inequality we have that

$2*(a^{2} + b^{2}) \ge (a + b)^{2} \Longrightarrow 2*(18 - c^{2}) \ge (6 - c)^{2}$

$\Longrightarrow 36 - 2c^{2} \ge 36 - 12c + c^{2} \Longrightarrow 0 \ge 3c^{2} - 12c$

$\Longrightarrow 0 \ge 3c(c - 4) \Longrightarrow 0 \le c \le 4.$

With $c = 0$ we have $a = b = 3$ and with $c = 4$ we have $a = b = 1.$ So we have established that the minimum and maximum values of $c$ are $0$ and $4,$ respectively.

Next, by the quadratic formula the roots of $x^{2} - (m + 2)x + 5m$ are

$x = \dfrac{(m + 2) \pm \sqrt{(m + 2)^{2} - 20m}}{2} = \dfrac{(m + 2) \pm \sqrt{(m - 8)^{2} - 60}}{2}.$

Now the first thing we can observe from this result is that, for $x$ to be real for integer $m,$ we require that $(m - 8)^{2} - 60 \ge 0,$ i.e., that either $m \le 0$ or $m \ge 16.$ Next, applying the " $c$ condition", we require that exactly one of these roots lies on the interval $(0,4).$ Now for $m \ge 16$ we know that both roots are positive and one is $\ge 9$ so only the choice of the $-$ sign could possibly yield a root in the desired interval. (Note that we know both roots are positive since clearly one is and the product of the roots, namely $5m,$ is positive.) So with a bit of algebraic manipulation, we find that we would require that

$0 \lt (10 + ((m - 8) - \sqrt{(m - 8)^{2} - 60}) \lt 8 \Longrightarrow 0 \lt (10 + A) \lt 8,$

where $A = (m - 8) - \sqrt{(m - 8)^{2} - 60}.$ But clearly $A \gt 0$ for $m \ge 16,$ and thus the inequality cannot be satisfied, i.e., there are no cases for $m \ge 16$ where there is exactly one root in the interval $(0,4).$

Now we need to look at the $m \le 0$ option. For $m = 0$ the given quadratic becomes $x^{2} - 2x = x(x - 2) = 0,$ which, with roots $0,2,$ does satisfy the condition that exactly one root lies in the interval $(0,4).$ For $m \lt 0$ we know that, since the product of the roots, namely $5m,$ is negative, one root will be positive and one negative. So the only chance that one of the roots will lie in the interval $(0,4)$ will be if we choose the $+$ sign in our equation for $x.$ Then with some algebraic manipulation, we find that we would require that

$(m + 2) + \sqrt{(m - 8)^{2} - 60} \lt 8$

$\Longrightarrow \sqrt{(m - 8)^{2} - 60} \lt -(m - 8) - 2$

$\Longrightarrow (m - 8)^{2} - 60 \lt (m - 8)^{2} + 4(m - 8) + 4$

$\Longrightarrow -64 \lt 4(m - 8) \Longrightarrow - 16 \lt (m - 8) \Longrightarrow -8 \lt m.$

Thus the integral values of $m$ that satisfy the given requirements are those between $-7$ and $0$ inclusive, giving us a total of $\boxed{8}$ integral values.