Find the equation of a line which passes through the point $(1, 1, 1)$ and intersect the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x + 2}{1} = \frac{y - 3}{2} = \frac{z + 1}{4}$

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Solution

$l_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$
$l_{2} : \frac{x + 2}{1} = \frac{y - 3}{2} = \frac{z + 1}{4}$
Let $2 r + 1, 3 r + 2, 4 r + 3$ be the point of intersection of $l_{1}$ and $l_{2}$
$∴ \frac{3 r + 2 - 3}{2} = \frac{2 r + 1 + 2}{1} = \frac{4 r + 3 + 1}{4} - - - (i)$
$r = - 2$
but $r$ does not satisfy equatton $(i)$
$∴$ $l_{1}$ and $l_{2}$ are skew lines.
points $(1, 2, 3)$ and $(- 2, 3, - 1)$ lies on $l_{1}$ and $l_{2}$
if three points are collinear $(x_{1} y_{1} z_{1}), (x_{2} y_{2} z_{2}) (x_{3} y_{3} z_{3})$ , then
$△ = ∣ ∣ ∣ ∣ \begin{matrix} x_{1} - x_{2} & y_{1} - y_{2} & z_{1} - z_{2} x_{2} - x_{3} & y_{2} - y_{3} & z_{2} - z_{3} x_{3} - x_{1} & y_{3} - y_{1} & z_{3} - z_{1} \end{matrix} ∣ ∣ ∣ ∣ = 0$
checking colinearity of $(1, 1, 1), (1, 2, 3), (- 2, 3, - 1)$
$△ = 0$ $∴$ they are collinear.
Hence equation of line is $\frac{(x - 1)}{- 3} = \frac{(y - 1)}{1} = \frac{(z - 1)}{- 4}$

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Find the equation of a line which passes through the point $(1, 1, 1)$ and intersect the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x + 2}{1} = \frac{y - 3}{2} = \frac{z + 1}{4}$