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 the $$xy$$-plane, then the value of $$a + b$$ is :

We parametrize the two lines. Line 1 passes through $$(-1, -a, -b-1)$$ with direction $$(3, 5, 7)$$, giving points $$(3t - 1,\; 5t - a,\; 7t - b - 1)$$. Line 2 passes through $$(2, b, 2a)$$ with direction $$(1, 4, 7)$$, giving points $$(s + 2,\; 4s + b,\; 7s + 2a)$$.

Setting coordinates equal at the intersection:

$$3t - 1 = s + 2 \implies s = 3t - 3 \quad \ldots (1)$$

$$5t - a = 4s + b \quad \ldots (2)$$

$$7t - b - 1 = 7s + 2a \quad \ldots (3)$$

Substituting $$s = 3t - 3$$ into equation (2):

$$5t - a = 12t - 12 + b \implies 7t = 12 - a - b \quad \ldots (2')$$

Substituting into equation (3):

$$7t - b - 1 = 21t - 21 + 2a \implies 14t = 20 - 2a - b \quad \ldots (3')$$

From $$(2')$$, $$14t = 24 - 2a - 2b$$. Equating with $$(3')$$:

$$24 - 2a - 2b = 20 - 2a - b \implies -b = -4 \implies b = 4$$

From $$(2')$$: $$7t = 12 - a - 4 = 8 - a$$, so $$t = \frac{8 - a}{7}$$.

Since the intersection point lies on the $$xy$$-plane, the $$z$$-coordinate is zero:

$$7t - b - 1 = 0 \implies 7t = 5 \implies t = \frac{5}{7}$$

So $$\frac{8 - a}{7} = \frac{5}{7}$$, giving $$a = 3$$.

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

Hence, the correct answer is Option 3.