Q).Find the equations of the straight line that is perpendicular to the lines (2x-4)/3 = (y-6)/1 = (3 -6z)/1 and (x-2)/-1 = (3-y)/-1 = (2z-4)/-4 and which passes through the midpoint of the points (2 , 6 , 1/2) and (2 , 3 , 2)???

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.

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....

A plane containing point \(\displaystyle (x_0, y_0, z_0)\) and having vector <A, B, C> as its normal vector has equation \(\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0\) (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.

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.