Largest solution

$\begin{aligned} \sqrt{3x^2-18x+52} + \sqrt{2x^2-12x+162} & = \sqrt{-x^2+6x+280} \\ \sqrt{3(x^2-6x+9)+25} + \sqrt{2(x^2-8x+9)+144} & = \sqrt{-(x^2-6x+9)+289} \\ \sqrt{3{\color{#3D99F6}(x-3)^2}+25} + \sqrt{2{\color{#3D99F6}(x-3)^2}+144} & = \sqrt{-{\color{#3D99F6}(x-3)^2}+289} & \small \color{#3D99F6} \text{Let }u = (x-3)^2 \\ \sqrt{3{\color{#3D99F6}u}+25} + \sqrt{2{\color{#3D99F6}u}+144} & = \sqrt{-{\color{#3D99F6}u}+289} & \small \color{#3D99F6} \text{Squaring both sides} \\ 3u+25 + 2\sqrt{(3u+25)(2u+144)} +2u+144 & = 289 - u & \small \color{#3D99F6} \text{Rearranging} \\ 2\sqrt{6u^2+482u+3600} & = 120 - 6u & \small \color{#3D99F6} \text{Squaring both sides} \\ 4(6u^2+482u+3600) & = 36u^2 -1440u + 14400 & \small \color{#3D99F6} \text{Rearranging} \\ 12u^2-3368u & = 0 \\ u(3u-842) & = 0 \end{aligned}$

$\implies \begin{cases} u = (x-3)^2 = 0 & \implies x = 3 \\ u = (x-3)^2 = \dfrac {842}3 & \implies x = \sqrt{\dfrac{842}3} + 3 \end{cases}$

Therefore, the integral solution is $\boxed{3}$ .