Find the equation of the plane given a line and a point

Question

I’m trying to solve a problem where i have to find the plane equation that contains a given straight line and a given point.

In this photo you can see the equation of the straight Line and the given point $P = (1,-2,3)$. The answer of the problem is the last Line. What i tried is this:

0. transform the straight line into parametric form
1. find the director parameters of the straight line $V = ( L,m,n)$
2. find a point of the straight line $Q = ( f, g, h)$
3. find the vector $PQ$.
4. do the vector product of the vectors $V$ and $PQ$ that gives me the director parameters of the plane that I'm searching.

The problem is that at this point what i find is that the director parameters of the plane are $N = ( -\frac{1}{3}, \frac{10}{3}, -\frac{14}{9})$ that are different from the solution ones.

Can anyone show me how to solve this problem so i can check and understand where I’m doing it wrong?

Answer 1

This is a real quick method:

Find the pencil of planes such that $H_{\alpha, \beta}=\alpha (2x-3y+z-3)+\beta (x+3y+2z+1)=0$
(all these are planes that have the line given by the exercise in common).

Then, you have to impose that $P=(1,-2,3) \in H_{\alpha, \beta}$.

This condition gives us $\alpha (2+6+3-3)+ \beta (1+6-6+1)=0$ $\implies$ $\beta= -4\alpha$.

If we now substitute $(\alpha , \beta)=(1,-4)$ in the equation of $H_{\alpha, \beta}$ we have $2x-3y+z-3-4x-12y-8z-4=0$, so

$$2x+15y+7z+7=0$$

Answer 2

Here is part of the solution using your method:

Answer 3

If you want another way (but a little informal) to do it, you can do this:
Give $z=1$, for example, then, you'll have the system
$\left\{ \begin{array}{lcc} 2x-3y=2\\ x+3y=-3\end{array}\right.$.
If you solve the sistem, you wil find a point that belongs to the line. Give $z=0$, then, the new system is
$\left\{ \begin{array}{lcc} 2x-3y=3\\ x+3y=-1\end{array}\right.$.
Newly, solve the system.
Thus, you have three points of the plane that you want. With them, you can find the equation of the plane.