<s>[INST] 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. 
 [/INST] 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.
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