Since all the numbers that accompany the letters a, b and c are even, then you can apply common-factor to factor the 2.
Then you have
[tex]\begin{gathered} 10a+6b+6c=2\cdot5a+2\cdot3b+2\cdot3c \\ 10a+6b+6c=2(5a+3b+3c) \end{gathered}[/tex]
Therefore, the given simplified expression is
[tex]2(5a+3b+3c)[/tex]
