The lines `x + y + z - 3 = 0 = 2x - y + 5z - 6 and x - y - z + 1 = 0 = 2x + 3y + 7z - k` are coplanar then k equals

The lines x+y+z-3=0=2x-y+5z-6 and x-y-z+1=0=2x+3y+7 are coplanar then k equals

Prove that the lines 3x + 2y + z – 5 = 0 = x + y – 2z – 3 and 2x – y – z = 0 = 7x + 10y – 8z – 15 are perpendicular

Add 2x - 3y + z , 5y - x + 7z and 3x - y - 6z.

If the lines 2x-y+3z + 4 = 0=ax + y-z + 2 and x-3y + z=0 =x + 2y + z +1 are coplannar then the value of a is ...........

The planes 2x + 5y + 3z = 0, x-y+4z = 2 and7y-5z + 4 = 0

The (x+4)/(3) = (y+6)/(5) = (z-1)/(-2) and 3x - 2y + z+5 =0 = 2x + 3y + 4z -k are coplanar for k is equal to

Solve the system: x + y - 2z = 0, 2x - 3y + z = 0, 3x - 7y + 10z = 0, 6x - 9y + 10z = 0.

If the planes 2x + 3y - z + 5 = 0, x + 2y – kz + 7 = 0 are perpendicular then k=