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

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$D' = (1 + pqr)D$

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(B) Using the properties 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 swapping columns twice to restore the original order $(b, c, a 	o a, b, c)$,we get:$\begin{vmatrix} b_1 & c_1 & a_1 \ b_2 & c_2 & a_2 \ b_3 & c_3 & a_3 \end{vmatrix} = (-1)^2 \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = D$Therefore,$D' = D + pqrD = (1 + pqr)D$.

If $D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ 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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