Question

Sol. [Key Concept: The intersection lies on $z = 0$; equate the two lines' general points and solve for $a + b$.]

Line $\left( L_{1} \right)$

$$\frac{x + 1}{3} = \frac{y + a}{5} = \frac{z + b + 1}{7} = r_{1} $$⇒ General point P on $L_{1}$ is

$$\left( 3r_{1} - 1,5r_{1} - a,7r_{1} - b - 1 \right) $$Line ( $L_{2}$ )

$$\frac{x - 2}{1} = \frac{y - b}{4} = \frac{z - 2a}{7} = r_{2} $$⇒ General point Q on $L_{2}$ is

$$\left( r_{2} + 2,4r_{2} + b,7r_{2} + 2a \right) $$For point of intersection

$$\begin{array}{r} 3r_{1} - 1 = r_{2} + 2 \Rightarrow r_{2} = 3r_{1} - 3\#(1) \end{array}$$

$$\begin{array}{r} 5r_{1} - a = 4r_{2} + b\#(2) \end{array}$$

$$\begin{array}{r} 7r_{1} - b - 1 = 7r_{2} + 2a\#(3) \end{array}$$

Since point lies on XY plane

$\Rightarrow z -$ coordinate $= 0$

From $L_{1}:7r_{1} - b - 1 = 0 \Rightarrow 7r_{1} = b + 1$

From $L_{2}:7r_{2} + 2a = 0 \Rightarrow 2a = - 7r_{2}$

Substitute $r_{2} = 3r_{1} - 3$

$${2a = - 7\left( 3r_{1} - 3 \right) }{\Rightarrow a = \frac{- 21r_{1} + 21}{2} }$$Put in equation (2)

$${5r_{1} - a = 4r_{2} + b }{5r_{1} - \left( \frac{- 21r_{1} + 21}{2} \right) = 4\left( 3r_{1} - 3 \right) + \left( 7r_{1} - 1 \right) }{\Rightarrow r_{1} = \frac{5}{7} }{\Rightarrow b = 7r_{1} - 1 = 4 }{\Rightarrow a = 3\therefore a + b = 7 }$$