Find the plane $P$ passing through the origin such that the three planes $P$, $P_1=(x+y+z=1)$ and $P_2= (x-y+z=2)$ meet along a line in R3. What I did is find the equation of a line of the intersection like
$$(x+y+z=1)+t(x-y+z=2)=0$$
Since it passes through the origin, I substitute $x=0$, $y=0$, $z=0$ into the equation and get $t$. Then I sub $t$ into the equation to obtain the equation of plane. 
 A: You’ve got the right idea, but there’s a detail that you might’ve overlooked. Every plane that passes through the intersection of $P_1$ and $P_2$ has an equation that’s a nontrivial linear combination of their equations: $$s(x+y+z-1)+t(x-y+z-2)=0.$$ What you’ve done is to set $s=1$ in this equation, which excludes $P_1$ from the set. For this problem, we know that $P_1$ doesn’t pass through the origin and therefore can’t be the solution, so it’s OK to do this, but you need to be careful that you don’t exclude a solution by doing this in other similar problems.  
Continuing from the above linear combination, setting $x=y=z=0$ reduces the equation to $-s-2t=0$, or $s=-2t$. (In fact, since the constant term of a plane equation that passes through the origin is zero, we could’ve reached this constraint without bothering to compute any of the other terms.) Taking $t=-1$ produces the equation $x+3y+z=0$, which clearly passes through the origin as required.
A: HINT
We can proceed as follows


*

*the general equation for $P$ is $ax+by+cz=0$ with normal $\vec n=(a,b,c)$

*find the common line to $P_1$ and $P_2$; note that it suffices to find a common point $Q$ and the direction vector that is $\vec v=\vec n_1\times \vec n_2$

*then $\vec  n=\vec {OQ}\times \vec v$
A: Adding x + y + z = 1 and x - y + z = 2 gives 2x + 2z = 3 or z = 3/2- x.  putting that into x+ y+ z= 1, x+ y+ 3/2- x= y+ 3/2= 1 so y= -3/2.  Taking t= x as parameter, the line of intersection is x= t, y= -3/2, z= 3/2- t.  Any plane that contains the origin can be written as z= Ax+ By.  In order that it contains that line we must have 3/2- t= At- (3/2)B= 0, so (A+ 1)t=(5/2)B  for all t.  In order that this be true for all t, we must have A- 1= 0 and (5/2)B= 0.  That is, z= x.
