If \((x - 1)\) and \((x - 3)\) are factors of the polynomial \(px^3 + qx^2 + 11x - 6\), then the polynomial must be divisible by both \((x - 1)\) and \((x - 3)\). 

This means that if we substitute \(x = 1\) and \(x = 3\) into the polynomial, the result must be equal to zero in both cases. Let's use this property to find the values of \(p\) and \(q\).

Substituting \(x = 1\) into the polynomial, we get \(p(1)^3 + q(1)^2 + 11(1) - 6 = p + q + 11 - 6 = p + q + 5\). Since the polynomial must be divisible by \((x - 1)\), this expression must be equal to zero. So, we have \(p + q + 5 = 0\).

Substituting \(x = 3\) into the polynomial, we get \(p(3)^3 + q(3)^2 + 11(3) - 6 = 27p + 9q + 33 - 6 = 27p + 9q + 27\). Since the polynomial must be divisible by \((x - 3)\), this expression must be equal to zero. So, we have \(27p + 9q + 27 = 0\).

Now, we have a system of two equations with two unknowns: \(p + q + 5 = 0\) and \(27p + 9q + 27 = 0\). Solving this system of equations, we find that \(p = \boldsymbol{1}\) and \(q = \boldsymbol{-6}\). 

So, the values of \(p\) and \(q\) that make \((x - 1)\) and \((x - 3)\) factors of \(px^3 + qx^2 + 11x - 6\) are \(p = \boldsymbol{1}\) and \(q = \boldsymbol{-6}\).