1. Base inductiva ($n=0$):
$$ \begin{aligned} 3^{3(0)+2} + 5 \cdot 2^{3(0)+1} &= 3^2 + 5 \cdot 2^1 \\ &= 9 + 10 = 19 \end{aligned} $$
Es divisible por $19$.

2. Hipótesis inductiva:

3. Paso inductivo ($n=k+1$):
Analizamos $3^{3(k+1)+2} + 5 \cdot 2^{3(k+1)+1}$:
$$ \begin{aligned} 3^{3k+3+2} + 5 \cdot 2^{3k+3+1} &= 3^3 \cdot 3^{3k+2} + 2^3 \cdot 5 \cdot 2^{3k+1} \\ &= 27(3^{3k+2}) + 8(5 \cdot 2^{3k+1}) \end{aligned} $$
Descomponemos el coeficiente $27$ como $19 + 8$:
$$ \begin{aligned} &= (19 + 8) \cdot 3^{3k+2} + 8(5 \cdot 2^{3k+1}) \\ &= 19(3^{3k+2}) + 8(3^{3k+2} + 5 \cdot 2^{3k+1}) \end{aligned} $$
Sustituyendo la hipótesis inductiva:
$$ \begin{aligned} &= 19(3^{3k+2}) + 8(19m) \\ &= 19(3^{3k+2} + 8m) \end{aligned} $$
La expresión resultante es claramente múltiplo de $19$.

$$ \boxed{(3^{3n+2} + 5 \times 2^{3n+1}) \text{ es divisible por } 19} $$