The plane \({{\pi }_{1}}\) contains the point \(A\left( 2,1,-1 \right)\) and the line \(l\) with equation \(\frac{x-5}{2}=y\), \(z=1\). Another plane \({{\pi }_{2}}\) meets \({{\pi }_{1}}\) in the line \(l\). It is known that \(\left( \begin{matrix}
-1 \\
2 \\
-6 \\
\end{matrix} \right)\) is a normal for \({{\pi }_{2}}\).
Show that the equation of \({{\pi }_{1}}\) in scalar product form is \(\mathbf{r}.\left( \begin{matrix}
-2 \\
4 \\
5 \\
\end{matrix} \right)=-5\).[3]
Find the position vector of the foot of perpendicular, \(F\) from \(A\) to the line \(l\).
Hence or otherwise, find the perpendicular distance from \(A\) to \({{\pi }_{2}}\).[3]
Find the acute angle between \({{\pi }_{1}}\) and \({{\pi }_{2}}\).[2]
\(A’\) is the point of reflection of \(A\) about the plane \({{\pi }_{2}}\). Deduce the area of the triangle \(AFA’\), leaving your answer in the form \(\frac{a\sqrt{5}}{b}\), where \(a\) and \(b\) are integers to be determined.[3]