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Expanding, gives us $\left(1+i\right)\left(1+2i\right)=-1+3i$. Then, expanding again gives us $\left(-1+3i\right)\left(1+3i\right)=\boxed{-10}.$

(1 + i)(1 + 2i)(1 + 3i)

= (1 + 2i - i) (1 + 2i + i) (1 + 2i)

= ((1 + 2i)^2 + 1)(1 + 2i)

= (-3 + 4i + 1)(1 + 2i)

= (-2 + 4i)(1 + 2i)

= -2(1 - 2i)(1 + 2i)

= -2 (5)

= -10