Answer
A plane that is perpendicular to the two planes has equation:
$ - x + y + z = d$, ${\ \ \ }$ where $d$ is any number.
Work Step by Step
By Theorem 1, the normal vector of the plane $x+y=3$ is ${{\bf{n}}_1} = \left( {1,1,0} \right)$, and the normal vector of the plane $x+2y-z=4$ is ${{\bf{n}}_2} = \left( {1,2, - 1} \right)$.
The plane required is perpendicular to the two planes $x+y=3$ and $x+2y-z=4$. Therefore, its normal vector is perpendicular to ${{\bf{n}}_1}$ and ${{\bf{n}}_2}$. Such normal vector ${\bf{n}}$ is obtained by taking the cross product of ${{\bf{n}}_1}$ and ${{\bf{n}}_2}$. Thus,
${\bf{n}} = {{\bf{n}}_1} \times {{\bf{n}}_2} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&1&0\\
1&2&{ - 1}
\end{array}} \right|$
${\bf{n}} = \left| {\begin{array}{*{20}{c}}
1&0\\
2&{ - 1}
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
1&0\\
1&{ - 1}
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
1&1\\
1&2
\end{array}} \right|{\bf{k}}$
${\bf{n}} = - {\bf{i}} + {\bf{j}} + {\bf{k}}$
So, the equation of the plane required is
${\bf{n}}\cdot\left( {x,y,z} \right) = d$
$\left( { - 1,1,1} \right)\cdot\left( {x,y,z} \right) = d$
$ - x + y + z = d$, ${\ \ \ }$ where $d$ is any number.
We may also write the equation of the plane as
$x - y - z = e$, ${\ \ \ }$ where $e$ is any number.
