Solution:

The answer is no. Note that $10!$ has exactly one prime factor of $7$. Therefore, exactly one of $a$, $b$, and $c$ is divisible by $7$. If $7$ divides $b$ (we write this as $7 \mid b$), then $7 \mid a b + b c$ but $7 \nmid a c$, so the equation cannot hold. Likewise, if $7 \mid a$, then $7 \mid a b$, $7 \mid a c$ but $7 \nmid b c$, and if $7 \mid c$, then $7 \mid b c$, $7 \mid a c$ but $7 \nmid a b$. So there are no solutions.