We use $x\oplus y$ for the bitwise XOR of $x$ and $y$.
Define the XOR-product of $x$ and $y$, denoted by $x\otimes y$, similar to a long multiplication in base $2$, except that the intermediate results are XORed instead of the usual integer addition.
For example, $7\otimes 3=9$, or in base $2$, ${111}_2\otimes {11}_2={1001}_2$: $$\begin{array}{r}\phantom{\otimes 111}{111}_2\\ \otimes \phantom{1111}{11}_2\\ \phantom{\otimes 111}{111}_2\\ \oplus \phantom{11}{111}_2\phantom{9}\\ \phantom{\otimes 11}{1001}_2\end{array}$$ We consider the equation: $$\begin{array}{r}(a\otimes a)\oplus (2\otimes a\otimes b)\oplus (b\otimes b)=c\otimes c\end{array}$$

For example, $(a,b,c)=(1,2,1)$ is a solution to this equation, and so is $(1,8,13)$.

Let $F(N)$ be the number of solutions to this equation satisfying $0\le a\le b\le N$. You are given $F(10)=21$.

Find $F({10}^7)$.