Answer
$x+y=1$
Work Step by Step
For Step 1:
The equations of transformation from cylindrical coordinates to rectangular coordinates are given by:
\begin{align*}
x &= r\cos\theta \\
y &= r\sin\theta \\
z &= z
\end{align*}
The equations of transformation from rectangular coordinates to cylindrical coordinates are given by:
\begin{align*}
r &= \sqrt{x^2 + y^2} \\
\theta &= \tan^{-1}\left(\frac{y}{x}\right) \\
z &= z
\end{align*}
For Step 2:
If $a,b,c$ and $\vec{d}$ are constants and $a\neq 0, b\neq 0, c\neq 0$, then the graph of the equation $ax+by+cz+d=0$ is a plane with normal vector $\vec{n}=\langle a,b,c\rangle$.
We notice that the plane passes through $P(1,0,0), Q(0,1,0)$, and $R(1,0,1)$. Therefore, the vectors $\vec{u}=\overrightarrow{PQ}=\langle -1,1,0\rangle$ and $\vec{v}=\overrightarrow{PR}=\langle 0,0,1\rangle$ are on the plane.
Since $\vec{u}$ and $\vec{v}$ are on the plane, their cross product $\vec{u}\times\vec{v}$ is normal to the plane. Hence,
\begin{align*}
\vec{u}\times\vec{v} &= \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-1 & 1 & 0 \\
0 & 0 & 1
\end{vmatrix} = \mathbf{i} + \mathbf{j} = \langle 1,1,0\rangle
\end{align*}
Thus, the equation of the plane can be written as:
\begin{align*}
1(x) + 1(y) + 0(z) &= d \
x + y &= d
\end{align*}
Since the plane passes through $P(1,0,0)$, we have $d=1$. Finally, the equation of the plane is:
\begin{align*}
x+y&=1
\end{align*}
