Using the property of determinants and without expanding, prove that:

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

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

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

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

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

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

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)

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

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

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

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 $\mathrm{R}_{1} \leftrightarrow \mathrm{R}_{2}$ and $\mathrm{R}_{2} \leftrightarrow \mathrm{R}_{3}$, we have:

$\Delta_{2}=(-1)^{2}\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|=\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$    ...(3) 

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.