Answer
All the planes parallel to the plane passing through the points $\left( {1,2,3} \right)$, $\left( {1,2,7} \right)$, and $\left( {1,1, - 3} \right)$ have equations $x=c$, where $c$ is any number.
Work Step by Step
Let $A = \left( {1,2,3} \right)$, $B = \left( {1,2,7} \right)$, and $C = \left( {1,1, - 3} \right)$ denote the points that the plane $\Pi$ passes through. Thus, we have the vectors:
$\overrightarrow {AB} = B - A = \left( {1,2,7} \right) - \left( {1,2,3} \right) = \left( {0,0,4} \right)$
$\overrightarrow {AC} = C - A = \left( {1,1, - 3} \right) - \left( {1,2,3} \right) = \left( {0, - 1, - 6} \right)$
Since the vectors $\overrightarrow {AB} $ and $\overrightarrow {AC} $ lie in the plane, their cross product ${\bf{n}} = \overrightarrow {AB} \times \overrightarrow {AC} $ is a vector normal to the plane $\Pi$. So,
${\bf{n}} = \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
0&0&4\\
0&{ - 1}&{ - 6}
\end{array}} \right|$
${\bf{n}} = \left| {\begin{array}{*{20}{c}}
0&4\\
{ - 1}&{ - 6}
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
0&4\\
0&{ - 6}
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
0&0\\
0&{ - 1}
\end{array}} \right|{\bf{k}}$
${\bf{n}} = 4{\bf{i}}$ ${\ \ }$ or ${\ \ }$ ${\bf{n}} = \left( {4,0,0} \right)$
Now, all the planes parallel to the plane $\Pi$ has normal vector ${\bf{n}} = 4{\bf{i}}$. Thus, by Eq. (5) of Theorem 1 (Section 13.5), all the planes parallel to the plane $\Pi$ have equations:
$4x + 0\cdot y + 0\cdot z = d$,
$4x = d$, ${\ \ }$ for some $d$.
We may write $x=c$, where $c$ is any number. Hence, all the planes parallel to the plane $\Pi$ have equations $x=c$, where $c$ is any number.
