My approach:

\(\left\{{}\begin{matrix}3x^2+6xy-3y^2=6\\2x^2+2xy+2y^2=6\end{matrix}\right.\)

It implies that

 \(\left(3x^2+6xy-3y^2\right)-\left(2x^2+2xy+2y^2\right)=6-6=0\)

  \(x^2+4xy-5y^2=0\)

    \(\left[{}\begin{matrix}x=y\\x=5y\end{matrix}\right.\)

Case 1: x=y, subtitute to (1) we have 

\(x^2=1\) \(\Leftrightarrow\) \(x=\pm1\)

Then we get 

Let \(x\in R.\) Prove that

                \(\sqrt{6+4\sqrt{2}\sin x}+\sqrt{3-2\sqrt{2}\cos x}\le2+\sqrt{2}\)

We have: \(f\left(2\right)=5.2+1=11\)

               \(f\left(-1\right)=5.\left(-1\right)+1=-4\)

then: \(a=f\left(2\right)-f\left(-1\right)=11-\left(-4\right)=15\)

Hence: \(g\left(1\right)=a.1+3=15.1+3=18\)

               \(f\left(a\right)=\sqrt{25a^2-30a+9}\)

               \(g\left(a\right)=a\)

We have the following:

        \(f\left(a\right)=g\left(a\right)+7\)

\(\Leftrightarrow\sqrt{25a^2-30a+9}=a+7\)

\(\Leftrightarrow\left\{{}\begin{matrix}25a^2-30a+9=\left(a+7\right)^2\\a+7\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\ge-7\\24a^2-44a-40=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ge-7\\\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=-\dfrac{2}{3}\end{matrix}\right.\)

Finally, there are two values of a.