Equation of plane containing a point and a line
Andrew Mclaughlin 
Find the equation of the plane containing the point $A(0,1,-1)$ and the line $(d) : \begin{cases} 2x - y + z + 1 = 0 \\ x + y + 1 = 0 \end{cases}$
Where should I start? I was thinking about writing the normal vectors for the line and make their cross product or something like that, but I don't really understand what's going on here. I would really appreciate some help.
$\endgroup$2 Answers
$\begingroup$A plane in $\mathbb{R}^{3}$ satisfies an equation of the type
\begin{equation} ax+by+cz+d=0. \tag{1} \end{equation}You can reduce this problem to that of finding a plane defined by 3 points. So just find 2 points on the line $ d $, using the given equations. For $x=0$ and $x=1$ you get, respectively, the points $B\left( 0,-1,-2\right) $ and $C\left( 1,-2,-5\right) $. To find 3 out of all the constants $a,b,c,d$ you need to substitute the coordinates of $A,B,C$ into $(1)$:\begin{eqnarray*} \left\{ \begin{array}{c} b-c+d=0 \\ -b-2c+d=0 \\ a-2b-5c+d=0 \end{array} \right. &\Leftrightarrow &\left\{ \begin{array}{c} b-c+d=0 \\ -3c+2d=0 \\ a-2b-5c+d=0 \end{array} \right. \Leftrightarrow \ldots &\Leftrightarrow &\left\{ \begin{array}{c} b+\frac{1}{2}c=0 \\ d=\frac{3}{2}c \\ a-2b-5c+\frac{3}{2}c=0, \end{array} \right. \end{eqnarray*}whose solution in terms of $c$ is $d=\frac{3}{2}c,a=\frac{5}{2}c,b=-\frac{1}{2}c$. So
$$ \frac{5}{2}cx-\frac{1}{2}cy+cz+\frac{3}{2}c=0 $$and, finally, dividing by $c/2$ we obtain the equation of the plane$$ 5x-y+2z+3=0.\tag{2} $$
Plot of plane $(2)$, line $d$ and points $A, B, C$
Hint:
The pencil of planes through the line with equations $2x-y+z+1=0,\;x+y+1=0$, can be described by the equation$$\lambda (2x-y+z+1)+\mu(x+y+1)=0,\qquad \lambda,\mu\in\mathbf R,$$where $\lambda,\mu$ are defined are to a constant factor, i.e.$$(2\lambda+\mu)c+(-\lambda+\mu)y+\lambda z+\lambda+\mu=0$$Write that the point $A$ satisfies this equation and deduce the value of the ratio $\lambda:\mu$.
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