Cheesy Equation

We note that $|x^2-(7+k^2)x+7k^2| \ge 0$ and $\sqrt{(x-3)(x-3k+2)} \ge 0$ , then $|x^2-(7+k^2)x+7k^2| + \sqrt{(x-3)(x-3k+2)} = 0$ , when simultaneously,

$\begin{cases} |x^2-(7+k^2)x+7k^2| = |(x-7)(x-k^2)| = 0 & ...(1) \\ \sqrt{(x-3)(x-3k+2)} = 0 & ...(2) \end{cases}$

For $(1) = 0 \quad \Rightarrow (x-7)(x-k^2) = 0$

$\Rightarrow \begin{cases} x = 7 & \Rightarrow (2): 4(7-3k+2) = 0 & \Rightarrow k = 3 \\ x = k^2 & \Rightarrow (2): (k^2-3)(k^2-3k+2) = 0 \\ & \quad \quad \quad \space (k^2-3)(k-1)(k-2) = 0 & \Rightarrow k = \pm \sqrt{3}, 1, 2 \end{cases}$

For $(2) = 0 \quad \Rightarrow (x-3)(x-3k+2) = 0$

$\Rightarrow \begin{cases} x = 3 & \Rightarrow (1): -4(3-k^2) = 0 & \Rightarrow k = \pm \sqrt{3} \\ x = 3k-2 & \Rightarrow (1): (3k-9)(3k-k^2+2) = 0 \\ & \quad \quad \quad \space (k-3)(k-1)(k-2) = 0 & \Rightarrow k = 3, 1, 2 \end{cases}$

Therefore, the sum of all real values of $k$ $= -\sqrt{3} + 1 + \sqrt{3} + 2 + 3 = \boxed{6}$