Equation of a plane through intersection of two and parallel to other

Question

I have got two planes $$x+y+z=1$$ and $$2x+3y-z+4=0$$ . I am required to find a plane by intersection of two and parallel to $x$ axis . I think the plane parallel to $x$ axis so Simply $P1+kP2=0$ hence $$x(1+2k)+y(1+3k)+z(1-k)-1+4k=0$$ and how to create a relation by which this plane is parallel to $x$ axis and with any other given plane ?

Answer 1

You are right in taking the required plane to be $P_1+k P_2 = 0$ (Intersection of $P_1$ and $P_2$). Since the required plane is parallel to the $x$-axis, the co-efficient of $x$ in it's equation should be $0$. $2k+1 = 0 \implies k = \frac{-1}{2}$. The required plane is: $$ \frac{-y}{2} + \frac{3z}{2} - 3 = 0 $$

BTW, the $z$ co-efficient is $1-k$ and not $1-4k$.