$\Delta=\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & z+y\end{array}\right|$

$=\left|\begin{array}{ccc}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a & p & x\end{array}\right|+\left|\begin{array}{ccc}b+c & q+r & y+z \\ c+a & r+p & z+x \\ b & q & y\end{array}\right|$

$=\Delta_{1}+\Delta_{2}(\text { say })…….(1)$

Now, $\Delta_{1}=\left|\begin{array}{ccc}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a & p & x\end{array}\right|$

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

$\Delta_{1}=\left|\begin{array}{lll}b & q & y \\ c & r & z \\ a & p & x\end{array}\right|$

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

$\Delta_{1}=(-1)^{2}\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|=\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|……(2)$

$\Delta_{2}=\left|\begin{array}{ccc}b+c & q+r & y+z \\ c+a & r+p & z+x \\ b & q & y\end{array}\right|$

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

$\Delta_{2}=\left|\begin{array}{ccc}c & r & z \\ c+a & r+p & z+x \\ b & q & y\end{array}\right|$

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

$\Delta_{2}=\left|\begin{array}{lll}c & r & z \\ a & p & x \\ b & q & y\end{array}\right|$

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

$\Delta_{2}=(-1)^{2}\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|=\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

From $(1),(2),$ and $(3),$ we have:

$\Delta=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

Hence, the given result is proved.