Question

If the lines $\frac{x}{1}=\frac{y}{2}=\frac{z}{3},\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}$ and $\frac{x-a}{3}=\frac{y-1}{2}=\frac{z-2}{b}$ are concurrent, then the value of $b-2a$ is equal to

• A. $5$
• B. $7$
• C. $3$
• D. $1$

Answer 1

The correct option is C $3$

Given lines:
$L_1:\frac{x}{1}=\frac{y}{2}=\frac{z}{3}{L}_2:\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}{L}_3:\frac{x-a}{3}=\frac{y-1}{2}=\frac{z-2}{b}$
Let a point on line $L_1$ be $(\lambda ,2\lambda ,3\lambda )$ and a point on line $L_2$ be $(3\mu +1,-\mu +2,4\mu +3)$
At point of intersection:
$\lambda =3\mu +1\cdots \left(i\right)2\lambda =-\mu +2\cdots \left(ii\right)3\lambda =4\mu +3\cdots \left(iii\right)$
Solving $\left(i\right)$ and $\left(ii\right),$ we get
$\lambda =1,\mu =0$
So, point of intersection is $(1,2,3)$ lie on line $L_3$ for concurrency
$\Rightarrow \frac{1-a}{3}=\frac{1}{2}=\frac{1}{b}$
$\Rightarrow b=2,a=-\frac{1}{2}$
$\therefore b-2a=3$