$${(3 + 6x)^n} = {3^n}{(1 + 2x)^n}$$

If T₉ is numerically greatest term

$$\therefore$$ $${T_8} \le {T_9} \le {T_{10}}$$

$${}^n{C_7}{3^{n - 7}}{(6x)^7} \le {}^n{C_8}{3^{n - 8}}{(6x)^8} \ge {}^n{C_9}{3^{n - 9}}{(6x)^9}$$

$$ \Rightarrow {{n!} \over {(n - 7)!7!}}9 \le {{n!} \over {(n - 8)!8!}}3\,.\,(6x) \ge {{n!} \over {(n - 9)!9!}}{(6x)^2}$$

$$ \Rightarrow \underbrace {{9 \over {(n - 7)(n - 8)}}}_{} \le \underbrace {{{18\left( {{3 \over 2}} \right)} \over {(n - 8)8}} \ge {{36} \over {9.8}}{9 \over 4}}_{}$$

$$72 \le 27(n - 7)$$ and $$27 \ge 9(n - 8)$$

$${{29} \over 3} \le n$$and $$n \le 11$$

$$\therefore$$ $${n_0} = 10$$

For $${(3 + 6x)^{10}}$$

$${T_{r + 1}} = {}^{10}{C_r}$$

$${3^{10 - r}}{(6x)^r}$$

For coeff. of x⁶

$$r = 6 \Rightarrow {}^{10}{C_6}{3^4}{.6^6}$$

For coeff. of x³

$$r = 3 \Rightarrow {}^{10}{C_3}{3^7}{.6^3}$$

$$\therefore$$ $$k = {{{}^{10}{C_6}} \over {{}^{10}{C_3}}}.{{{3^4}{{.6}^6}} \over {{3^7}{{.6}^3}}} = {{10!7!3!} \over {6!4!10!}}.8$$

$$ \Rightarrow k = 14$$

$$\therefore$$ $$k + {n_0} = 24$$