Show that the line with parametric equations x = 6 + 8t, y = −5 + t, z = 2 + 3t does not intersect the plane with equation 2x − y − 5z − 2 = 0.

Answer 1

well let's just substitute ...

2(6+8t) - (-5+t) - 5(2+3t) = 0
12 + 16t + 5 -t - 10 - 15t = 0
7 = 0

This was false when I went to school , and I don't think it has changed since then.

So, since we end up with a contradiction, (7=0), there is no solution, and the line does not intersect with the plane.

or

direction of the line is <8,1,3>
normal to the plane is <2, -1, -5>

<8,1,3> dot <2, -1, -5>
= 16 -1 - 15 = 0

thus the line and the normal are perpendicular, or
the line is parallel to the plane.
However, my second method would not be complete, since our line could lie on the plane itself. So we still have to perform the calculation of my first method to show it does NOT lie on the plane