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

Let the point of intersection be $(x, y, 0)$ since it lies on the $xy$-plane.

For the first line, equating the coordinates to a parameter $\lambda$:

$\dfrac{x+1}{3} = \dfrac{y+a}{5} = \dfrac{0+b+1}{7} = \lambda$

This gives $x = \dfrac{3b+3}{7} - 1$ and $y = \dfrac{5b+5}{7} - a$.

For the second line, equating the coordinates to a parameter $\mu$:

$\dfrac{x-2}{1} = \dfrac{y-b}{4} = \dfrac{0-2a}{7} = \mu$

This gives $x = \dfrac{-2a}{7} + 2$ and $y = \dfrac{-8a}{7} + b$.

Equating the $x$-coordinates from both lines:

$\dfrac{3b+3}{7} - 1 = \dfrac{-2a}{7} + 2$

$\dfrac{3b-4}{7} = \dfrac{-2a+14}{7}$

$2a + 3b = 18$

Equating the $y$-coordinates from both lines:

$\dfrac{5b+5}{7} - a = \dfrac{-8a}{7} + b$

$5b + 5 - 7a = -8a + 7b$

$a - 2b = -5$

Solving the two linear equations $2a + 3b = 18$ and $a - 2b = -5$:

From the second equation, $a = 2b - 5$. Substituting this into the first equation:

$2(2b - 5) + 3b = 18$

$4b - 10 + 3b = 18$

$7b = 28 \Rightarrow b = 4$

Substituting $b = 4$ into $a = 2b - 5$:

$a = 2(4) - 5 = 3$

Therefore, $a + b = 3 + 4 = 7$.

Answer: $7$