I need to find the vector equation for the line $t$ that is concurrent to both:
$$r:X = (1,1,-1)+\lambda(2,1,-1)$$ and $$s:\\x+y-3z = 1\\2x-y-2z=0$$
And also, $t$ is parallel to $MN$ when: $$M = (1,-1,4), N = (0,3,-1)\implies\vec {MN} = (-1,-2,-5)$$
I tried to write a generic point for $r$:
$$P = (1+2\lambda, 1+\lambda, -1-\lambda)$$ And in $s$ when we let $x = \lambda$ we get the generic point:
$$Q = (\lambda, 1+\frac{4}{5}\lambda, \frac{3}{5}\lambda)$$
I tried to calculate $$\vec {PQ} = (-1+t-2\lambda, \frac{4}{5}t-\lambda, 1+\frac{3t}{5} + \lambda)$$
And then make $$\vec {PQ} = a\vec{MN}$$ for some alpha. When I did this, I got no solution. Please help me :(
Taking $P=(1+2\lambda,1+\lambda,-1-\lambda)$ as above,
we can represent the general point on s by $Q=(2+5t,2+4t,1+3t)$.
Then $\vec{PQ}=\langle1+5t-2\lambda,1+4t-\lambda,2+3t+\lambda\rangle$, so
setting $\vec{PQ}=k(\vec{MN})=k\langle -1,4,-5\rangle$ gives the system of equations
$\;\;\;5t-2\lambda+k=-1$, $\;\;\;4t-\lambda-4k=-1$, $\;\;\;3t+\lambda+5k=-2$.
Now solve for $t$ and $\lambda$ to find the points $P$ and $Q$, and then you can write a vector equation for the line.