Find the line that intersects two other lines

The problem is as following : Find the line that contains the point A(-4,-5,3) and intersects the lines:

p: x=3t-1 , y=-2t-3 , z=-t+2 and the line q: x=2t+2 , y=2t-1 , z=-5t+1

I have been dealing with a good amount of problems with lines and planes in 3D, but I have no idea how to approach this one. I have no idea how to get the direction vector that's missing or just another point on the line I have to find. Any help would be appreciated. Thanks in prior.

Solution 1:

$L_P:\overrightarrow r = - \hat i - 3\hat j + 2\hat k + s\left( {3\hat i - 2\hat j - \hat k} \right) = \vec c + s\vec b$

$L_Q:\overrightarrow r = 2\hat i - \hat j + \hat k + t\left( {2\hat i + 2\hat j - 5\hat k} \right) = \vec a + t\vec d$

$L_R:\overrightarrow r = - 4\hat i - 5\hat j + 3\hat k + u\left( {a\hat i + b\hat j + c\hat k} \right) = \vec e + u\vec f$

$\vec f \times \vec d = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ a&b&c\\ 2&2&{ - 5} \end{array}} \right| = - \left( {5b + 2c} \right)\hat i + \left( {5a + 2c} \right)\hat j + \left( {2a - 2b} \right)\hat k \& \vec e - \vec a = - 6\hat i - 4\hat j + 2\hat k$

$\left( {\vec e - \vec a} \right).\left( {\vec f \times \vec d} \right) = 0 \Rightarrow 6\left( {5b + 2c} \right) - 4\left( {5a + 2c} \right) + 2\left( {2a - 2b} \right) = 0$

$30b + 12c - 20a - 8c + 4a - 4b = 0 \Rightarrow - 16a + 26b + 4c = 0...(i)$

$\vec f \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ a&b&c\\ 3&{ - 2}&{ - 1} \end{array}} \right| = \left( { - b + 2c} \right)\hat i + \left( {a + 3c} \right)\hat j - \left( {2a + 3b} \right)\hat k$

$\vec e - \vec c = - 3\hat i - 2\hat j + \hat k$

$\left( {\vec e - \vec c} \right).\left( {\vec f \times \vec b} \right) = 0 \Rightarrow - 3\left( { - b + 2c} \right) - 2\left( {a + 3c} \right) - \left( {2a + 3b} \right) = 0$

$3b - 6c - 2a - 6c - 2a - 3b = 0 \Rightarrow - 4a - 12c = 0...(ii)$

$c = k;a = - 3k$

$ \Rightarrow - 16a + 26b + 4c = 0 \Rightarrow - 16a \times \left( { - 3k} \right) + 26b + 4k = 0 \Rightarrow b = - \frac{{52k}}{{26}} \Rightarrow b = - 2k$

$\left\langle {a,b,c} \right\rangle = \left\langle { - 3, - 2,1} \right\rangle $

$L_R:\overrightarrow r = - 4\hat i - 5\hat j + 3\hat k + u\left( { - 3\hat i - 2\hat j + \hat k} \right) \Rightarrow {L_R}:\frac{{x + 4}}{{ - 3}} = \frac{{y + 5}}{{ - 2}} = \frac{{z - 3}}{1}$

This is the final equation