Levi-Civita and Kronecker Delta

Question

To prove this, $$ \sum_{pq} \epsilon_{ipq} \epsilon_{jpq} = 2\delta_{ij} $$

I used Levi-Civita and delta relation

$$ \sum_{q} \epsilon_{ipq}\epsilon_{jpq} = \delta_{ij}\delta_{pp}-\delta_{ip}\delta_{pj} $$ then first and second term both will be just $\delta_{ij}$ so both cancel out.

Where it is going wrong? Since using that relation I am getting zero not as twice of delta.

Answer 1

If the indices run from $1$ to $3$ then $$\sum_p\delta_{pp}=3$$ which gives the desired result.