TITLE: Find an equation of the plane.
QUESTION [0 upvotes]: Find an equation of the plane.
The plane that passes through the line of intersection of the planes 
$x − z = 3$ and $y + 4z = 2$
 and is perpendicular to the plane 
$x + y − 3z = 5$
I did the cross product of the first two vectors and then cross product of that vector with the third vector but not sure if I am doing it correctly.

REPLY [1 votes]: A plane passing through the intersection of planes $H_1=0$ and $H_2=0$ belongs to the pencil generated by these planes, i.e. it has  equation $\lambda H_1+\mu H_2=0$ for some $\lambda, \mu$.
In the present case, the equation will be
$$\lambda(x-z)+\mu(y+4z)=\lambda x +\mu y+ (4\mu-\lambda)z=3\lambda+2\mu, $$
 and it will be perpendicular to the plane $x-y+3z=5$ if they have orthogonal normal vectors.
As a vector normal to the plane is $(\lambda, \mu, 4\mu-\lambda)$, the orthogonality condition is 
$$\lambda-\mu+3(4\mu-\lambda)=-2\lambda+11\mu=0.$$
A solution (defined within a non-zero factor) is $\lambda=11$, $\mu=2$, whence the equation
$$ 11x+2y-3z=37. $$
{"set_name": "stack_exchange", "score": 0, "question_id": 1933311}