Easy part- the midpoint of (2, 6, 1/2) and (2, 3, 2) is ((2+2)/2, (6+3)/2, (1/2+ 2)/2)= (2, 9/2, 5/4).

A vector pointing in the direction of the first line is <3, 1, 1> and a vector pointing in the diredction of the second line is <-1, -1, -4>. A vector perpendicular to both is their cross product, <-3, 11, -2>. The line you want goes in that direction and contains (2, 9/2, 5/4): (x- 2)/-3= (y- 9/2)/11= z+ 4/-2.

A plane containing point and having vector <A, B, C> as its normal vector has equation (I bet you knew that!) However, there are an infinite number of vectors normal to a given line and there are an infinite number of planes containing that particular point and parallel to that particular line.Q) Find the equation of the plane that passes through the point (1,-2,3) and that is parallel to the line (x-2)/3 = (y-1)/2 = (z+5)/-2?

please help me in solving these problems....

Imagine a line through the given point parallel to that line. Now, any plane containing this new line satisfies the conditions. There isn't enough information to give a specific answer.