Answer: 925

Small Hint:

Write the three arithmetic-sequence terms as $x,$ $x+r,$ and $x+2r$

Big Hint:

Subtract adjacent square equations, then subtract those two resulting equations

Solution:

Let the three terms be $x,$ $x+r,$ and $x+2r.$ Subtracting the square equations gives $$r(2x+r)=300-36=264$$ and $$r(2x+3r)=596-300=296.$$ Their difference is $2r^2=32,$ so $r=\pm4.$ Negating all three terms does not change their squares, so take $r=4.$ Then $4(2x+4)=264,$ giving $x=31.$ Therefore $k=x^2-36=961-36=925.$