Find the direction ratio of the line `(x-1)/2=3y=(2z+3)/4`

The distance of the point (2,3,-4) from the point of intersection of the line (x-1)/-3 = (y-2)/2 = (z-2)/1 and the plane 2x-3y-z-20 =0 is

Find the angle between the line (x+1)/(2)=(y)/(3)=(z-3)/(6) and the plane 10x+2y-11z=3

Find the angle between the line (x+1)/2=y/3=(z-3)/6 and the plane 10x + 2y - 11z = 3.

Find the angle between the line (x+1)/(2)= (y)/(3)= (z-3)/(6) and the plane 10x+2y-11z=3

The point of intersection of the line (x-5)/3 = (y-7)/(-1) = (z+2)/1 , (x+3)/(-36) = (y-3)/2 = (z-6)/4 is

If the distance between the plane Ax+2y+z=d and the plane containing the lines (x-1)/2 = (y-2)/3 = (z-3)/4 and (x-2)/3 = (y-3)/4 = (z-4)/5 is sqrt6 , then |d| is equal to

The distance of the point (1,2,1) from the line (x-1)/2=(y-2)/1=(z-3)/2 is

Find the angle between the following pairs of lines : (x-2)/2=(y-1)/5=(z+3)/(-3) " and " (x+2)/(-1)=(y-4)/8=(z-5)/4 .

The distance of the point (3,8,2) from the line (x-1)/2 = (y-3)/4 = (z-2)/3 measured parallel to the plane 3x+2y-2z +15 = 0 is