If all the λs are different, then the eigenvectors of A are necessarily orthogonal and the cs are eigenvalues (assuming the $b$s are also normalized). If some of the λs are the same, this is not necessarily the case.
As an example what goes wrong when some $\lambda$ are equal, take the case $A= I_2$ (a $3\times 3$ example is also easy to construct). We have
$$ A= \begin{pmatrix} b_1 & b_2 \end{pmatrix}
\begin{pmatrix} c_1 & 0 \\\ 0 & c_2 \end{pmatrix}
\begin{pmatrix} b_1^T \\\ b_2^T \end{pmatrix} $$
with
$$
\begin{align}
c_1 &=1 & c_2 &=2 \\
b_1 &= \frac{1}{\sqrt{2}}\begin{pmatrix} 1\\1\end{pmatrix} &
b_2 & = \frac{1}{2} \begin{pmatrix} 1 \\ -1 \end{pmatrix}
\end{align},
$$
so the $b$s are orthogonal but the $c$s are not the eigenvalues. (Calle post shows that in this case the $b$s have to be orthogonal)
To find an example where the $b$s are not orthogonal, take $A=0_2$. We can write the matrix in the form given above with
$$
\begin{align}
c_1 &=1 & c_2 &=-1 \\
b_1 &= \begin{pmatrix} 1\\1\end{pmatrix} &
b_2 & = -b_1
\end{align},
$$
so neither are the $c$s the eigenvalues nor the $b$s orthogonal.
For the case when all the $\lambda$s are different, the $b$s are up to normalization uniquely determined. They are automatically orthogonal on each other. If they are not normalized the $c$s are not necessarily the eigenvalues. But if you assume them normalized then the $c$s correspond to the $\lambda$s.