Answer
$y+z=1$
Work Step by Step
Theorem
If $a,b,c$ and $d$ are constants and $a,b,c$ are not all zero, then the graph of the equation
\begin{align}
ax+by+cz+d=0
\end{align}
is a plane has vector $\mathbf{n}=\langle a,b,c\rangle$ as a normal.
From the figure we notice that the plane passe through $P(0,0,1)$, $Q(0,1,0)$ and $R(1,1,0)$, then,
\begin{align}
\mathbf{u}=\overrightarrow{PQ}=\langle0,1,-1\rangle,\quad \mathbf{v}=\overrightarrow{QR}=\langle1,0,0\rangle
\end{align}
are vectors on the plane. Therefore, $\mathbf{n}=\mathbf{u}\times\mathbf{v}$ is a normal vector to the plane which equation can be written as
\begin{align}
ax+by+cz=d
\end{align}
where $\mathbf{n}=\langle a,b,c\rangle$. Since
\begin{align}
\mathbf{n}=\mathbf{u}\times\mathbf{v}=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 1 & -1 \\
1 & 0 & 0 \
\end{vmatrix}=0\mathbf{i}-\mathbf{j}-\mathbf{k}=\langle0,-1,-1\rangle
\end{align}
the equation results:
\begin{align}
\left(x\right)-\left(y\right)-\left(z\right)=d=-y-z=d
\end{align}
Since the plane passes through $P(0,0,1)$ we have that
\begin{align}
-0-1=d\Rightarrow d=-1
\end{align}
Thus, the equation of the plane is:
\begin{align}
y+z=1
\end{align}
