Using properties of determinants, prove that:

$\Delta=\left|\begin{array}{lll}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-2 R_{1}$ and $R_{3} \rightarrow R_{3}-3 R_{1}$, we have:

$\Delta=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 3 & 7+3 p\end{array}\right|$

Applying $R_{3} \rightarrow R_{3}-3 R_{2}$, we have:

$\Delta=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 0 & 1\end{array}\right|$

Expanding along $\mathrm{C}_{1}$, we have:

$\Delta=1\left|\begin{array}{cc}1 & 2+p \\ 0 & 1\end{array}\right|=1(1-0)=1$

Hence, the given result is proved.