Formula used:(a - b)2 = a2 + b2 - 2abCalculation:Given:\(?_1 = \begin{vmatrix} 1 & p & q \\ 1 & q & r \\ 1 & r & p\end{vmatrix} \ \text{and} \ ?_2 = \begin{vmatrix} 1 & 1 & 1 \\ q & r & p \\ r & p & q \end{vmatrix} \)Now, \(?_1 = \begin{vmatrix} 1 & p & q \\ 1 & q & r \\ 1 & r & p\end{vmatrix} \)R2 ? R2 - R1 and R3 ? R3 - R1? \(?_1 = \begin{vmatrix} 1 & p & q \\ 0 & q-p & r-q \\ 0 & r-p & p-q\end{vmatrix} \)? ?1 = 1.{(q - p)(p - q) - (r - p)(r - q)} ? ?1 = - (p - q)2 - (r - p)(r - q)? ?1 = - {p2 - 2pq + q2} - {(r - p)(r - q)}? ?1 = - p2 + 2pq - q2} - {r2 - qr - pr + pq}? ?1 = - p2 + 2pq - q2 - r2 + qr + pr - pq? ?1 = - p2 - q2 - r2 + qr + pr + pq ------(i)Again, \(?_2 = \begin{vmatrix} 1 & 1 & 1 \\ q & r & p \\ r & p & q \end{vmatrix} \)C2? C2 - C1 and C3 ? C3 - C1? \(?_2 = \begin{vmatrix} 1 & 0 & 0 \\ q & r-q & p-q \\ r & p-r & q-r \end{vmatrix} \)? ?2 = 1.{(r - q)(q - r) - (p - q)(p - r)}? ?2 = - (q - r)2 - (p - q)(p - r)? ?2 = -{q2 + r2 - 2qr} - {p2 - pq - pr + qr}? ?2 = - q2 - r2 - p2 + pq + pr + qr ------(ii)Adding equations (i) and (ii), we get,?1 + ?2 = - p2 - q2 - r2 + qr + pr + pq - q2 - r2 - p2 + pq + pr + qr ? ?1 + ?2 = -(p2 + q2 - 2pq + q2 + r2 - 2qr + r2 + p2 - 2pr)? ?1 + ?2 = -[(p - q)2 + (q - r)2 + (r - p)2] which is always negative.?1 + ?2 is always negative.