If $p$ and $q$ are zeroes of the polynomial $p(y) = 21y^2 - y - 2$, then find the value of $(1 - p) \cdot (1 - q)$.

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Step 1: $p + q = \frac{1}{21}$, $pq = \frac{-2}{21}$
Step 2: $(1 - p)(1 - q) = 1 - (p + q) + pq = 1 - \frac{1}{21} + \left(\frac{-2}{21}\right) = 1 - \frac{3}{21} = 1 - \frac{1}{7} = \frac{6}{7}$
Step 3: $(1 - p)(1 - q) = \frac{18}{21} = \frac{6}{7}$

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