Difficulties to understand the color factor calculation for the colour singlet state

Question

I have some difficulties to understand how the color factor

$$f = \frac{1}{4} \sum\limits_{a=1}^8 \langle c_1 ¦ \lambda_a ¦ c_3 \rangle \langle c_2 ¦ \lambda_a ¦ c_4 \rangle$$

are calculated generally and in particular for the singlet state $\frac{1}{\sqrt{3}}(r\bar{r}) + b\bar{b})  + g\bar{g})$.

The $\lambda_a$ are the Gell-Mann matrices, the $¦c_i>$ can be represented as column vectors

$¦r \rangle = (1,0,0)^T$, $¦b \rangle = (0,1,0)^T$, $¦g \rangle = (0,0,1)^T$ whereas the $\langle c_i ¦$ can be represented by the corresponding row vectors.

Looking at the definition, I thought that to obtain the result for the color singlet state ($f = \frac{4}{3}$ one just has to add the color factors for $r\bar{r}$, $b\bar{b}$, $g\bar{g}$ states together and multiply this by $\frac{1}{\sqrt{3}}$.

For the $r\bar{r}$ state  I assumed $c_1 = c_3 = (c_2 = c_4) = (1,0,0)^T which resulted in

$f_{r\bar{r}} = \frac{1}{4}(\lambda^3_{11} \lambda^3_{11} + \lambda^8_{11} \lambda^3_{11}) = \frac{1}{3}$
 

Applying these considerations, I obtained $f_{b\bar{b}} = \frac{1}{3}$ and $f_{g\bar{g}} = \frac{1}{3}$ which finally resulted in $f = \frac{1}{\sqrt{3}}$ for the color singlet state which is wrong.

There are obviously things I did not understand correctly...