Using properties of determinants,prove that:
$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$

(A) Let $\Delta = \left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|$.
Applying the row operations $R_{2} \rightarrow R_{2}-2 R_{1}$ and $R_{3} \rightarrow R_{3}-3 R_{1}$:
$\Delta = \left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2-2(1) & (3+2p)-2(1+p) & (4+3p+2q)-2(1+p+q) \\ 3-3(1) & (6+3p)-3(1+p) & (10+6p+3q)-3(1+p+q)\end{array}\right|$
$\Delta = \left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 3 & 7+3p\end{array}\right|$.
Now,applying the row operation $R_{3} \rightarrow R_{3}-3 R_{2}$:
$\Delta = \left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 3-3(1) & (7+3p)-3(2+p)\end{array}\right|$
$\Delta = \left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 0 & 1\end{array}\right|$.
Expanding along the first column $(C_{1})$:
$\Delta = 1 \cdot \left|\begin{array}{cc}1 & 2+p \\ 0 & 1\end{array}\right| - 0 + 0 = 1(1 - 0) = 1$.
Hence,the determinant is equal to $1$.