Which of the following planes intersects the planes $x-y+2z =3$ and $4x+3y-z=1$ along the same line?

Question

Which of the following planes intersects the planes $x-y+2z =3$ and $4x+3y-z=1$ along the same line?

$a)\space 11x+10y-5z=0$

$b)\space 7x+7y-4z=0$

$c)\space 5x+2y+z=2$

Answer 1

We have $x=3+y-2z$ putting this in the second equation we get $$4(3+y-2z)+3y-z=1 \\ 7y-9z=-11$$

Take $z=t$ and we get equation of line as $$(\frac{-5t+10}{7},\frac{9t-11}{7}, t)$$

$$11\times\frac{-5t+10}{7}+10\times\frac{9t-11}{7}- 5t=\frac{-55t+110+90t-110-35t} {7}=0$$ Hence it is the first option