Using properties of determinants, prove that \begin{vmatrix} b+c & c+a & a+b \ p+q & q+r & r+p \ x+y & y+z & z+x \end{vmatrix} = 2 \begin{vmatrix} a & b & c \ p & q & r \ x & y & z \end{vmatrix}

Let C1, C2, C3 be first, second and third columns respectively
\begin{vmatrix} b+c & c+a & a+b \ p+q & q+r & r+p \ x+y & y+z & z+x \end{vmatrix}
Applying C1 = C1 + C2 + C3, we get,
= 2 \begin{vmatrix} a+b+c & c+a & a+b \ p+q+r & q+r & r+p \ x+y+z & y+z & z+x \end{vmatrix}
Applying C2 = C2 - C1 and C3 = C3 - C1, we get,
= 2 \begin{vmatrix} a+b+c & -b & -c \ p+q+r & -p & -q \ x+y+z & -x & -y \end{vmatrix}
Finally, Applying C1 = C1 - (C2 + C3), we get,
= 2 \begin{vmatrix} a & b & c \ p & q & r \ x & y & z \end{vmatrix}
Hence, proved!