To find the coefficient of \(x^3\) in the expansion of \((1 - 3x)^6 + (1 + ax)^5\), we consider the contributions from each binomial expansion separately.

For \((1 - 3x)^6\), the term containing \(x^3\) is given by the binomial coefficient:

\(\binom{6}{3} (1)^{6-3} (-3x)^3 = 20(-27x^3) = -540x^3.\)

For \((1 + ax)^5\), the term containing \(x^3\) is:

\(\binom{5}{3} (1)^{5-3} (ax)^3 = 10a^3x^3.\)

The total coefficient of \(x^3\) is:

\(-540 + 10a^3 = 100.\)

Solving for \(a\), we have:

\(10a^3 = 640\)

\(a^3 = 64\)

\(a = 4.\)