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(D) Using the property of determinants,we can split the determinant $D'$ into the sum of determinants.$D' = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b

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$D' = D(1 + pqr)$

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(D) Using the property of determinants,we can split the determinant $D'$ into the sum of determinants.$D' = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} + \begin{vmatrix} pb_1 & qc_1 & ra_1 \ pb_2 & qc_2 & ra_2 \ pb_3 & qc_3 & ra_3 \end{vmatrix}$$D' = D + pqr \begin{vmatrix} b_1 & c_1 & a_1 \ b_2 & c_2 & a_2 \ b_3 & c_3 & a_3 \end{vmatrix}$By applying column swaps,$\begin{vmatrix} b_1 & c_1 & a_1 \ b_2 & c_2 & a_2 \ b_3 & c_3 & a_3 \end{vmatrix} = - \begin{vmatrix} a_1 & c_1 & b_1 \ a_2 & c_2 & b_2 \ a_3 & c_3 & b_3 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = D$.Thus,$D' = D + pqrD = D(1 + pqr)$.

Suppose $D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ and $D' = \begin{vmatrix} a_1 + pb_1 & b_1 + qc_1 & c_1 + ra_1 \\ a_2 + pb_2 & b_2 + qc_2 & c_2 + ra_2 \\ a_3 + pb_3 & b_3 + qc_3 & c_3 + ra_3 \end{vmatrix}$,then

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