For the following systems of linear equation,
$$
x_1+x_2+x_3=3
$$
$$
2x_1+3x_2-4x_3=3
$$
$$
-3x_1-4x_2+(k^2-6)x_3=2k
$$
Suppose that the solution set forms a line $\ell$. Find a scalar equation for the plane $P$ that contains the point (2,1,1) and is perpendicular to $\ell$.
My solution:
The augmented matrix is
$$
\begin{bmatrix}
1 & 0 & 7 & 6\\
0 &1&-6&-3\\
0 & 0 & k^2-9 & 2k+6
\end{bmatrix}
$$
For $k=3$, there is no solution.
For $k^2-9\neq 0$, there is a unique solution.
For $k=-3$, there is the infinite solution.
So for solution set forms a line $\ell$, $k=-3$. Then augmented matrix is
$$
\begin{bmatrix}
1 & 0 & 7 & 6\\
0 &1&-6&-3\\
0 & 0 &0 & 0
\end{bmatrix}
$$
The general solution is
$$
\{x=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}=\begin{bmatrix}
6 \\
-3 \\
0
\end{bmatrix}+c\begin{bmatrix}
-7 \\
6 \\
1
\end{bmatrix}, \, c\in R\}
$$
But how to find a $P$?
Best Answer
You already obtained the equation of the line.
$l: (6, -3, 0) + (-7, 6, 1) t$
You are seeking equation of the plane that is perpendicular to the line and passes through point $(2, 1, 1)$. As the plane is perpendicular to the line, its normal vector must be in the direction of the line. So, equation of the plane is:
$- 7 (x_1 - 2) + 6 (x_2 - 1) + (x_3 - 1) = 0$