If the line `(x+1)/3=(y-2)/4=(z+6)/5` is parallel to the planes `2x-3y+kz=0`, then the value of k is

If the line (x-1)/2=(y+3)/1=(z-5)/(-1) is parallel to the plane px + 3y - z + 5 = 0 , then the value of p -

If the line (x-1)/2=(y+3)/1=(z-5)/(-1) is parallel to the plane px + 3y - z + 5 = 0 , then the value of p -

IF the line (x-1)/2=(y+1)/3=z/4 and origin lie on the plane 4x+4y-kz=0, then k=

IF the line (x-1)/2=(y+1)/3=z/4 and origin lie on the plane 4x+4y-kz=0 , then k=

Find the value of k for which the line (x - 2)/3 = (1 - y)/k = (z - 1)/4 is parallel to the plane 2x + 6y + 3z - 4 = 0.

if an angle between the line,and the plane, (x-1)/2=(y-2)/1=(z-3)/-2 and the plane x-2y-kz= 3 is cos^-1 (2sqrt2)/3 then a value of k is:

IF the line (x-1)/2=(y+1)/3=z/4 and origin lie on the plane 4x+4y-kz=0, then k=? a)1 b)3 c)5 d)7