Intersection of line and plane

1. The problem statement, all variables and given/known data

Find the equation of the plane which passes through the line [itex]\frac{x-2}{3}=\frac{y-3}{1}=\frac{z+1}{2}[/itex]
and it is normal to the plane x+4yy-3z+7=0

2. Relevant equations



3. The attempt at a solution

I find the intersection point of the plane and the line. It is M(2,3,-1). Also I got the condition (A,B,C)(3,1,2)=0

[tex]\left\{\begin{matrix}
2A+3B-C+D=0 & \\
A+4B-3C=0 &
\end{matrix}\right.[/tex]

What is the third condition ?

 

Sorry I must fix something. Up in there is x+4y-3z+7=0, and (A,B,C)(1,4,-3)=0.

 

Is (A,B,C)(3,1,2)=0 the third condition?