Question
If the distances of the point $(1,2,a)$ from the line
$\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines
$L_{1}:\frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b}$ and
$L_{2}:\frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal,
then $a + b + c$ is equal to
Options
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 $A(3\lambda + 1,4\lambda + 2,\text{ }b\lambda + a)$
It lies on L
$$\therefore\frac{3\lambda}{1} = \frac{4\lambda + 2}{2} = \frac{b\lambda + a - 1}{1} $$$\Rightarrow \lambda = 1$ and $a + b - 1 = 3$
$$\begin{array}{r} \Rightarrow A(4,6,4),a + b = 4\#(1) \end{array}$$
Let $B(\mu + 1,4\mu + 2,c\mu + a)$
It also lies on L
$${\frac{\mu}{1} = \frac{4\mu + 2}{2} = \frac{c\mu + a - 1}{1} }{\Rightarrow 2\mu = 4\mu + 2 }{\Rightarrow \mu = - 1 }{a - c - 1 = - 1 }{\Rightarrow a = c }{\ldots(2)\&\text{ }B(0, - 2,0) }$$also $PA = PB,P(1,2,a),A(4,6,4)$
$${\Rightarrow 9 + 16 + (a - 4)^{2} = 1 + 16 + a^{2} }{\Rightarrow 16 + 8 = 8a }{a = 3\ \therefore c = 3,\text{ }b = 1 }{\therefore a + b + c = 7}$$
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