I have calculated (a) to be $(1,-2,2)^t, (-2,1,2)^t, (2,2,1)^t$. For (b) I have made all of these of unit length ie taken 1/3 of each vector. I have verified these are orthonormal by checking $<v1,v2>=<v1,v3>=<v2,v3>=0$. I have then computed $det[v1,v2,v3]$ and found it not equal to zero, so vectors are LI, matrix is of rank 3, so $v1,v2,v3$ form an orthonormal basis. This completes (b).
I am however stuggling with (c). I have said $I=C^t*[v1,v2,v3]C$ but don't know where to go from here. Any help would be great!
Assuming that $B$ is defined to be the transformation of $A$ into the new coordinates (this needs to be clarified as the question doesn't say what $B$ is meant to be), if you take $C=\frac13[v1,v2,v3]$ then we have $$C=\frac13\begin{pmatrix}1&-2&2\\-2&1&2\\2&2&1\end{pmatrix}$$ and so $C=C^t$ and $CC=I$.\begin{align}B=C^t AC&=\frac19\begin{pmatrix}1&-2&2\\-2&1&2\\2&2&1\end{pmatrix}\begin{pmatrix}1&-2&2\\-2&1&2\\2&2&1\end{pmatrix}\begin{pmatrix}1&-2&2\\-2&1&2\\2&2&1\end{pmatrix} \\ &=\frac19\begin{pmatrix}1&-2&2\\-2&1&2\\2&2&1\end{pmatrix}\begin{pmatrix}5&-4&-2\\-10&2&-2\\10&4&-1\end{pmatrix} \\&=\frac19\begin{pmatrix}45&0&0\\0&18&0\\0&0&-9\end{pmatrix} \\&=\begin{pmatrix}5&0&0\\0&2&0\\0&0&-1\end{pmatrix}.\end{align}
We can see that this is not an inner product, as the vector $(1,1,\sqrt7)$ in the new basis would then have norm $0$.