Find Value of a plus b for Line Intersection on XY Plane

Let $P(x_0, y_0, z_0)$ be the point of intersection of the two given lines. Since $P$ lies on the $xy$-plane, its $z$-coordinate must be equal to zero, i.e., $z_0 = 0$.

The given equations of the lines are: $L_1: \frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} = \lambda$ $L_2: \frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} = \mu$

Since $P(x_0, y_0, 0)$ lies on $L_1$, we can express its coordinates in terms of the parameter $\lambda$: $x_0 = 3\lambda - 1$ $y_0 = 5\lambda - a$ $0 = 7\lambda - (b+1) \implies \lambda = \frac{b+1}{7}$

Similarly, since $P(x_0, y_0, 0)$ lies on $L_2$, we can express its coordinates in terms of the parameter $\mu$: $x_0 = \mu + 2$ $y_0 = 4\mu + b$ $0 = 7\mu + 2a \implies \mu = -\frac{2a}{7}$

Now, equating the expressions for the $x$-coordinate from both lines: $3\lambda - 1 = \mu + 2$ $3\left(\frac{b+1}{7}\right) - 1 = -\frac{2a}{7} + 2$

Multiply the entire equation by $7$: $3(b+1) - 7 = -2a + 14$ $3b - 4 = -2a + 14$ $2a + 3b = 18 \quad \text{--- (Equation 1)}$

Next, equating the expressions for the $y$-coordinate from both lines: $5\lambda - a = 4\mu + b$ $5\left(\frac{b+1}{7}\right) - a = 4\left(-\frac{2a}{7}\right) + b$

Multiply the entire equation by $7$: $5(b+1) - 7a = -8a + 7b$ $5b + 5 - 7a = -8a + 7b$ $a - 2b = -5 \quad \text{--- (Equation 2)}$

We now solve Equations (1) and (2) simultaneously. From Equation (2), we get: $a = 2b - 5$

Substitute $a = 2b - 5$ into Equation (1): $2(2b - 5) + 3b = 18$ $4b - 10 + 3b = 18$ $7b = 28 \implies b = 4$

Substituting $b = 4$ back into the expression for $a$: $a = 2(4) - 5 = 3$

Thus, the value of $a + b$ is: $a + b = 3 + 4 = 7$