The d.c.s. of the line `2x+1=3-2y=4z+5` are

The d.C.s of the line 3x-1=2y-3 = 4z-1 are

If the distance between the plane Ax 2y + z = d and the plane containing the lines 2 1x = 3 2y = 4 3z and 3 2x = 4 3y = 5 4z is 6 , then |d| is

If the distance between the plane Ax 2y + z = d and the plane containing the lines 2 1x = 3 2y = 4 3z and 3 2x = 4 3y = 5 4z is 6 , then |d| is

The image of the line (x -1)/(3) = (y -3)/(1) = (z -4)/(-5) in the plane 2x - y + z + 3=0 is the line-

The equation of line is (2x-5)/4=(y+4)/3=(6-z)/6 then the d.c.'s ofline parallel of this line are :

Find the S.D. between the lines : x/2=y/-3=z/1 and (x-2)/3=(y-1)/-5=(z+2)/2

If the equation of the line AB is (3 - x)/1=(y + 2)/-2=(z - 5)/4 then write the d.r.s of the line parallel to AB.

The image of the line (x-1)/3=(y-3)/1=(z-4)/(-5) in the plane 2x-y+z+3=0 is the line (1) (x+3)/3=(y-5)/1=(z-2)/(-5) (2) (x+3)/(-3)=(y-5)/(-1)=(z+2)/5 (3) (x-3)/3=(y+5)/1=(z-2)/(-5) (3) (x-3)/(-3)=(y+5)/(-1)=(z-2)/5

The image of the line (x-1)/3=(y-3)/1=(z-4)/(-5) in the plane 2x-y+z+3=0 is the line

The image of the line (x-1)/3=(y-3)/1=(z-4)/(-5) in the plane : 2x-y+z+3=0 is the line :