If $A$ and $B$ are the points $(3, 4, 5)$ and $(-1, 3, -7)$ respectively,find the equation of the set of points $P$ such that $PA^{2} + PB^{2} = k^{2}$,where $k$ is a constant.

Let the coordinates of point $P$ be $(x, y, z)$.
Given $A = (3, 4, 5)$ and $B = (-1, 3, -7)$.
Using the distance formula,$PA^{2} = (x-3)^{2} + (y-4)^{2} + (z-5)^{2} = x^{2} - 6x + 9 + y^{2} - 8y + 16 + z^{2} - 10z + 25 = x^{2} + y^{2} + z^{2} - 6x - 8y - 10z + 50$.
Similarly,$PB^{2} = (x+1)^{2} + (y-3)^{2} + (z+7)^{2} = x^{2} + 2x + 1 + y^{2} - 6y + 9 + z^{2} + 14z + 49 = x^{2} + y^{2} + z^{2} + 2x - 6y + 14z + 59$.
Given the condition $PA^{2} + PB^{2} = k^{2}$,we have:
$(x^{2} + y^{2} + z^{2} - 6x - 8y - 10z + 50) + (x^{2} + y^{2} + z^{2} + 2x - 6y + 14z + 59) = k^{2}$.
Combining like terms:
$2x^{2} + 2y^{2} + 2z^{2} - 4x - 14y + 4z + 109 = k^{2}$.
Dividing by $2$:
$x^{2} + y^{2} + z^{2} - 2x - 7y + 2z + \frac{109}{2} = \frac{k^{2}}{2}$.
Thus,the equation is $x^{2} + y^{2} + z^{2} - 2x - 7y + 2z = \frac{k^{2} - 109}{2}$.