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

L.H.S. $=$ $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{1 + p}&{1 + p + q}\\2&{3 + 2p}&{4 + 3p + 2q}\\3&{6 + 3p}&{10 + 6p + 3q}\end{array}} \right|$

Applying ${R_2} \to {R_2} - 2{R_1}$ and ${R_3} \to {R_3} - 3{R_1}$ ,

we get
L.H.S. $=$ $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{1 + p}&{1 + p + q}\\0&1&{2 + p}\\0&3&{7 + 3p}\end{array}} \right|$

Expanding along ${C_1}$,

we get
$= 1(7 + 3p - 6 - 3p) = 1(1) = 1 =$ R.H.S.
Hence, proved.