The configuration of three cells must come in a 45-45-90 triangle. There are two cases, both shown above: the triangle has legs parallel to the axes, or it has its hypotenuse parallel to an axis. The first case can be solved by noticing that each selection of four cells in the shape of a square corresponds to four such possibilities. There are $7^{2}$ possible squares of size $2 \times 2,6^{2}$ possible squares of size $3 \times 3$, and so on. The total for this first case is thus $4\left(7^{2}+6^{2}+\cdots+1^{2}\right)=560$. The second case can also be done by casework: each triangle in this case can be completed into an $n+1$ by $2 n+1$ rectangle, of which there are $7 \cdot 6+6 \cdot 4+5 \cdot 2$ (for $n=1,2,3$ respectively). Multiply this by 4 to get all orientations of the triangle. The final answer is $560+4(7 \cdot 6+6 \cdot 4+5 \cdot 2)=864$.