Two planes given
$$x-y+z=5 , \hspace{0.5cm}x+y+z=3 $$
Their intersection is a line $l$.Find the equation of a plane such that the line $l$ is perpendicular to that required plane and this plane passes through the $(0,0,0).$
My attempt:I solve the system $$x-y+z=5, \hspace{0.5cm}x+y+z=3 $$ and i get $x+z=4$ which is normal of the required plane so the equation of required plane must be $$x+z+d=0$$ after substituting $(0,0,0)$ i get $\hspace{0.2cm}x+z=0$.
So my answer is $$\hspace{0.2cm}x+z=0$$
Am i correct or not?If you have some other method then please tell me.
Thanks
Note that $l$ is in the direction of the cross product of the normal of the first two planes.
$$\begin{align}\text{Direction} &= \left(\begin{array}{c}1\\-1\\1\end{array}\right)\times\left(\begin{array}{c}1\\1\\1\end{array}\right)\\ &= \left(\begin{array}{c}-2\\0\\2\end{array}\right)\\ &\equiv\left(\begin{array}{c}-1\\0\\1\end{array}\right)\end{align}$$
Since the third plane passes through $(0, 0, 0)$ and is perpendicular to $l$, we can define the third plane as
$$\left(\vec{r} - \left(\begin{array}{c}0\\0\\0\end{array}\right)\right)\cdot\left(\begin{array}{c}-1\\0\\1\end{array}\right) = 0$$ $$\vec{r} \cdot\left(\begin{array}{c}-1\\0\\1\end{array}\right) = 0$$
In Cartesian form, this is simply $$-x + z = 0$$