The variable line $l_1$ passes through the point $A$ $(2, 1, -1)$ and is parallel to the direction $ti + j +(1-t)k$
The variable line $l_2$ passes through the point B(1, -2, 3) and is parallel to the direction $-2i +tj +(t+2)k$
The points $P$ on $l_1$ and $Q$ on $l_2$ are such that $PQ$ is perpendicular to both $l_1$ and $l_2$.
$(i)$ for all values of t the length of $PQ$ is $2\sqrt3$
$(ii)$ There is no value for of $t$ for which the planes $APB$ and $APQ$ are perpendicular to each other.
$(iii)$ The perpendicular distance from the point $A$ to the plane $BPQ$ is denoted by $p$. Show that $(27-2p^2)t^2 +(36-4p^2)t + 12 - 8p^2=0$
I was able to do part $(i)$ by first calculating the vector product of the direction vectors of both the lines, and then using the vector product as the normal and taking the point $A$, calculate the equation of plane. Then I used point B in the point to plane perpendicular distance formula to show the given result. I have one confusion in part one. What does is the significance (geometric or otherwise) of the phrase 'for all values of t'. For part $(ii)$ I have an idea about where to start i.e. I have guessed that the solution must come from here:- that when I dot the normals of $APB$ and $APQ$ and set them equal to zero, I will get an equation in $t$ that has no real roots. However this is just an idea and I was unable to move forward with it. I was no idea how to do part three. Any help would be appreciated