# Find an equation for the plane

Here is the whole question.

Find an equation of the plane that passes through the points $P(1,0,-1)$ and $Q(2,1,0)$ and is parallel to the line of intersection of the planes $x+y+z=5$ and $x+y-z=1$.

And the answer by the book is $7x-5y-2z=9$.

As far as I know, the general strategy for finding an equation of a plane is to construct it via its normal vector.

And for the problem, I thought the intersection information could be used this way : Since the intersection line is parallel to the plane, the orthogonal vector of the intersection vector would be normal vector for the plane in question.

I'm not sure this strategy could work, and even if it works, how given information can be used in this process.

• You can find a normal vector for the plane by taking the cross-product of the vector from P to Q with a vector parallel to the line of intersection of the two planes. I believe the book has the incorrect answer, since the plane they give intersects the line of intersection of the planes. (I think it should be $x+y-2z=3$.) – user84413 Jul 9 '14 at 14:45
• $4y+3z=11$ is a plane, not a line. And the other direction vector is just $Q-P$. – rogerl Jul 9 '14 at 15:00