Question

The equation of the line which passes through the point (1, 1, 1) and intersecting the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ is

• A. $\frac{x-1}{4}=\frac{y-1}{11}=\frac{z-1}{13}$
• B. $\frac{x-1}{17}=\frac{y-1}{-3}=\frac{z-1}{17}$
• C. $\frac{x-1}{13}=\frac{y-1}{5}=\frac{z-1}{-2}$
• D. $\frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}$

Answer 1

The correct option is D

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

Any line through the point (1,1,1) is
$\frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}..................\left(1\right)$

This line intersects the line
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow a:b:c\ne 2:3:4and\left|\begin{array}{ccc}1-1& 2-1& 3-1\\ a& b& c\\ 2& 3& 4\end{array}\right|=0\Rightarrow 0-1(4a-2c)+2(3a-2b)=0\Rightarrow a-2b+c=\mathrm{0..............}\left(2\right)$
Again the line (1) intersects the line
$\frac{x-(-2)}{1}=\frac{y-3}{2}=\frac{z-(-1)}{4}\Rightarrow a:b:c\ne 1:2:4and\left|\begin{array}{ccc}-2-1& 3-1& -1-1\\ a& b& c\\ 1& 2& 4\end{array}\right|\Rightarrow -3(4b-2c)-2(4a-c)-2(2a-b)=0\Rightarrow 6a+5b-4c=\mathrm{0...............}\left(3\right)$
From (2) and (3) by cross multiplication, we have
$\frac{a}{8-5}=\frac{b}{6+4}=\frac{c}{5+12}\Rightarrow \frac{a}{3}=\frac{b}{10}=\frac{c}{17}$

So, the required line is
$\frac{x-1}{3}=\frac{y-1}{10}=\frac{z-1}{17}$