Answer
(a) $(1,0,-6)$, $(0,3,-1)$, $1$, $\frac{1}{2}$
(b) See explanations.
Work Step by Step
(a) When $t=0$, we can get a point on Line 1 as $(1,0,-6)$, similarly for $t=1$, the point is $(0,3,-1)$. We can confirm these points are also on Line 2 by plug-in one of the coordinates into the Line 2 equation to get the corresponding t-value, e.g. Line 2 $x=-1+2t=1$, we get $t=1$ which gives the first point; and $x=-1+2t=0$, we get $t=\frac{1}{2}$ which gives the second point.
(b) Let $t=0$, we can get a point on Line 3 as $(0,3,-5)$. Use this point to test on Line 4, if it is on Line 4, we have $8-2t=0$ and $t=4$, the other values are then $y=-9+3\times4=3$ and $z=6-4=2$ which gives a point $(0,3,2)$ on Line 4, the different z-value shows that the test point on Line 3 is not on Line4, this means that these two lines are not the same.
