(i) The integers between 7 and 21 that are multiples of 5 are 10, 15, and 20. There are 3 such numbers out of 15 total numbers (7 to 21 inclusive), so the probability \(p\) that a randomly chosen integer is a multiple of 5 is \(\frac{3}{15} = 0.2\). Therefore, \(X \sim \text{Bin}(12, 0.2)\).

(ii) We need to calculate \(P(3 \leq X \leq 5)\). This is given by:

\(P(X = 3, 4, 5) = \binom{12}{3} (0.2)^3 (0.8)^9 + \binom{12}{4} (0.2)^4 (0.8)^8 + \binom{12}{5} (0.2)^5 (0.8)^7\)

Calculating each term:

\(\binom{12}{3} (0.2)^3 (0.8)^9 = 0.23622\)

\(\binom{12}{4} (0.2)^4 (0.8)^8 = 0.13287\)

\(\binom{12}{5} (0.2)^5 (0.8)^7 = 0.05315\)

Adding these probabilities gives \(0.23622 + 0.13287 + 0.05315 = 0.422\).

(iii) We want \(P(X = 0) < 0.01\). For \(X \sim \text{Bin}(n, 0.2)\), \(P(X = 0) = (0.8)^n\). We solve:

\((0.8)^n < 0.01\)

Taking logarithms, \(n \log(0.8) < \log(0.01)\).

\(n > \frac{\log(0.01)}{\log(0.8)} \approx 20.65\)

Thus, the smallest integer \(n\) is 21.