Answer
\[
0=6 z-6-x+7 y
\]
Work Step by Step
The equation of a plane passing through the point $(a, b, c)$ and having normal vector $\langle l, m, n\rangle$ is
\[
0=(-a+x)l+(-b+y)m+(-c+z)n
\]
The equation of the plane is
\[
\begin{array}{c}
0=(1+x) \cdot-1+(1+y) \cdot 7+(-2+z) \cdot 6 \\
0=6 z-6-x+7 y
\end{array}
\]
