Find $N$ ?

$No.\quad of\quad diagonals={ _{ N }{ C }_{ 2 } }-N=2N\\ N(N-1)/2=3N\\ N-1=6\\ N=7\\$

No. of diagnols in an n-sided polygon = ${\phantom{0}}^{n}{\text{C}}_2 - n$
$\begin{aligned} {\phantom{0}}^{n}{\text{C}}_2 - n & = 2n\\ \dfrac{n!}{(n - 2)! × 2!} & = 3n\\ \\ \dfrac{n(n - 1)(n - 2)!}{(n - 2)!} & = 6n\\ \\ n(n - 1) & = 6n\\ n^2 - n & = 6n\\ n^2 & = 7n\\ n & = \color{#3D99F6}{\boxed{7}} \end{aligned}$

No. of diagonals = $\dfrac{n(n-3)}{2}$ ,

$\rightarrow$ $2n$ $=$ $\dfrac{n(n-3)}{2}$

$\rightarrow$ $n - 3$ $=$ $4$

$\rightarrow \boxed{n = 7}$ .

$\text{Please feel free to comment \& correct me if I am wrong , Thank you.}$