More generally, let $(x,y,z)=\left(\sqrt5,\sqrt6,\sqrt7\right)$ so that $\left(x^2,y^2,z^2\right)=(5,6,7).$
We rewrite the original expression in terms of $x,y,$ and $z,$ then apply the difference of squares repeatedly:
$$\begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \left[((x+y)+z)((x+y)-z)\right]\left[((z+(x-y))(z-(x-y))\right] \\ &= \left[(x+y)^2-z^2\right]\left[z^2 - (x-y)^2\right] \\ &= \left[x^2+2xy+y^2-z^2\right]\left[z^2-x^2+2xy-y^2\right] \\ &= \left[2xy + \left(x^2+y^2-z^2\right)\right]\left[2xy - \left(x^2+y^2-z^2\right)\right] \\ &= \left(2xy\right)^2 - \left(x^2+y^2-z^2\right)^2 \\ &= \left(2\cdot\sqrt5\cdot\sqrt6\right)^2 - \left(5+6-7\right)^2 \\ &= \boxed{104}. \end{align*}$$
Remark
From this solution, note that the original expression has cyclic symmetry with respect to $x,y,$ and $z:$
$$\begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \cdots \\ &= \left(2xy\right)^2 - \left(x^2+y^2-z^2\right)^2 \\ &= 4x^2y^2 - x^4 - y^4 - z^4 - 2x^2y^2 + 2y^2z^2 + 2z^2x^2 \\ &= 2x^2y^2 + 2y^2z^2 + 2z^2x^2 - x^4 - y^4 - z^4. \end{align*}$$
~MRENTHUSIASM