Solution

$ L : \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1} $

$ L_1 : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b} = \lambda $

$ L_2 : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c} = \mu $

Let $ A(3\lambda + 1,; 4\lambda + 2,; b\lambda + a) $

It lies on $ L $

$ \Rightarrow \frac{3\lambda}{1} = \frac{4\lambda + 2}{2} = \frac{b\lambda + a - 1}{1} $

$ \Rightarrow \lambda = 1 $ and $ a + b - 1 = 3 $

$ \Rightarrow a + b = 4 \quad ...(1)$

$ A(4, 6, 4),; a + b = 4 $

Let $ B(\mu + 1,; 4\mu + 2,; c\mu + a) $

It also lies on $ L $

$ \Rightarrow \frac{\mu}{1} = \frac{4\mu + 2}{2} = \frac{c\mu + a - 1}{1} $

$ \Rightarrow 2\mu = 4\mu + 2 \Rightarrow \mu = -1 $

$ a - c - 1 = -1 \Rightarrow a = c \quad ...(2)$

Thus, $ a + b + c = 4 + c = 4 + a $