Right answer is (a) -1(1+m+n+q)
For explanation: Given that, Δ=\(\begin{vmatrix}1+m&n&q\\m&1+n&q\\n&m&1+q\end{vmatrix}\)
Applying C1→C1+C2+C3
Δ=\(\begin{vmatrix}1+m+n+q&n&q\\1+m+n+q&1+n&q\\1+m+n+q&m&1+q\end{vmatrix}\)=(1+m+n+q)\(\begin{vmatrix}1&n&q\\1&1+n&q\\1&m&1+q\end{vmatrix}\)
Applying R1→R2-R1
Δ=(1+m+n+q)\(\begin{vmatrix}0&1&0\\1&1+n&q\\1&m&1+q\end{vmatrix}\)
Expanding along the first row, we get
Δ=(1+m+n+q)(0-1(1+q-q)+0)
Δ=-1(1+m+n+q).