Consider the plane P in R-3 given by x-y-2z=0

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I found the matrix A whose columns are a basis for P, A=[1,-1,-2] (vertical form).

Using that I was able to find the projection matrix:

P=$\frac{-1}{2} \left( \begin{array}{cc} 1 & -1 & 2 \\ -1 & 1 &1 \\-2 &2 &-4 \end{array} \right)$

How would I find a vector e that is orthogonal to P and the projection matrix $Q=\frac{ee^T}{e^Te}$ and how is that related to P?