If the point of intersection of the lines \( \frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} \) and \( \frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} \) lies on \( xy \)-plane, then the value of \( a+b \) is:
\( 2 \)
\( 5 \)
\( 7 \)
\( 9 \)