Equation of the plane passing through the point and containing the line I should find an equation of the plane passing through the point 
$A=(1,2,-1)$ 
and containing the line:

2x - y + 5z + 3= 0;
x - 5y + 2z -1 =0;

So the line seems to have two equation, and I don't know how to find a single equation of the line. How can I describe the line or just find the plane? Should I transform the equations into one, or..? 

Hint:
Multiplying the 2nd equation by 2 and subtracting gives $z=-5-9y$, and then substituting back into either equation gives $x=23y+11$.
Therefore the line has parametric equations $x=11+23t, \;y=t,\; z=-5-9t$, so
$\vec{a}=\langle23,1,-9\rangle$ is a direction vector for the line.
Since $B=(11,0,-5)$ is a point on the line, $\vec{b}=\vec{AB}=\langle10,-2,-4\rangle$ is parallel to the plane.
Therefore $\vec{n}=\vec{a}\times\vec{b}$ will give a normal vector for the plane.
