Find the equation of the line which intersect the lines (x+2)/1 = (y-…

The correct answer is Option (1) → $\frac{x-1}{3} = \frac{y-1}{10} = \frac{z-1}{17}$ ##

The lines are

$\frac{x+2}{1} = \frac{y-3}{2} = \frac{z+1}{4} \dots (i)$

$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \dots (ii)$

Let $P(\lambda - 2, 2\lambda + 3, 4\lambda - 1)$ be any point on line (i) and $Q(2\mu + 1, 3\mu + 2, 4\mu + 3)$ be any point on line (ii). Also, the given point is $A(1, 1, 1)$.

For some definite value of $\lambda$ and $\mu$, the required line passes through $A, P$ and $Q$.

The direction ratios of $AP$ are $\lambda - 3, 2\lambda + 2, 4\lambda - 2$

The direction ratios of $AQ$ are $2\mu, 3\mu + 1, 4\mu + 2$

$∴\frac{\lambda - 3}{2\mu} = \frac{2\lambda + 2}{3\mu + 1} = \frac{4\lambda - 2}{4\mu + 2}$

$\Rightarrow \frac{\lambda - 3}{2\mu} = \frac{2\lambda + 2}{3\mu + 1} = \frac{2\lambda - 1}{2\mu + 1} = k \text{ (let)}$

$\Rightarrow \lambda - 3 = 2\mu k, 2\lambda + 2 = 3\mu k + k, 2\lambda - 1 = 2\mu k + k$

$\mu k = \frac{\lambda - 3}{2}, 2\lambda + 2 = 3\left(\frac{\lambda - 3}{2}\right) + k, 2\lambda - 1 = 2\left(\frac{\lambda - 3}{2}\right) + k$

$\Rightarrow 2\lambda - 1 = \lambda - 3 + k \Rightarrow k = \lambda + 2$

$\Rightarrow \mu k = \frac{\lambda - 3}{2}, k = \frac{\lambda + 13}{2}, k = \lambda + 2$

$∴\frac{\lambda + 13}{2} = \lambda + 2 \Rightarrow \lambda = 9$

Also $k = \lambda + 2 = 11$

Hence, the direction ratios of $AP$ are $6, 20, 34$ i.e., $3, 10, 17$

Therefore, equation of required line is

$\frac{x-1}{3} = \frac{y-1}{10} = \frac{z-1}{17}$