Radicals

$\begin{aligned} X & = \color{#3D99F6}\left(\sqrt 5 + \sqrt 6 + \sqrt 7\right) \left(\sqrt 5 + \sqrt 6 - \sqrt 7\right) \color{#D61F06} \left(\sqrt 5 - \sqrt 6 + \sqrt 7\right) \left(-\sqrt 5 + \sqrt 6 + \sqrt 7\right) & \small \color{#3D99F6} \text{Note that }(a+b)(a-b) = a^2-b^2 \\ & = \color{#3D99F6}\left((\sqrt 5 + \sqrt 6)^2 - 7\right) \color{#D61F06}\left(7 - (\sqrt 6 - \sqrt 5)^2 \right) \\ & = {\color{#3D99F6} 7 \left(\sqrt 6 + \sqrt 5 \right)^2 + 7 \left(\sqrt 6 - \sqrt 5 \right)^2} - \left(\sqrt 6 + \sqrt 5 \right)^2 \left(\sqrt 6 - \sqrt 5 \right)^2 - 49 \\ & = {\color{#3D99F6} 7 \left( 2 (6 +5) \right)} - \left(1 \right)^2 - 49 \\ & = \boxed{104} \end{aligned}$

The product of the first and second term is $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})=(\sqrt{5}+\sqrt{6})^2-(\sqrt{7})^2=5+2\sqrt{30}+6-7=4+2\sqrt{30}$

The product of the third and fourth term is

$(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})=(\sqrt{7}+(\sqrt{5}-\sqrt{6}))(\sqrt{7}-(\sqrt{5}-\sqrt{6})=(\sqrt{7})^2-(\sqrt{5}-\sqrt{6})^2=7-(5-2\sqrt{30}+6)=-4+2\sqrt{30}$

The final product then is

$(4+2\sqrt{30})(-4+2\sqrt{30})=-16+4(30)=104$