The value of $\begin{vmatrix}1&bc&bc(b+c)\\1&ca&ca(c+a)\\1&ab&ab(a+b)\end{vmatrix}$ is

0

The correct answer is Option (2) → 0

Given determinant:

$D = \begin{vmatrix} 1 & bc & bc(b+c) \\ 1 & ca & ca(a+b) \\ 1 & ab & ab(a+b) \end{vmatrix}$

Subtract first row from second and third rows (R2-R1, R3-R1):

$D = \begin{vmatrix} 1 & bc & bc(b+c) \\ 0 & ca-bc & ca(a+b) - bc(b+c) \\ 0 & ab-bc & ab(a+b) - bc(b+c) \end{vmatrix}$

Expand along first column:

$D = 1 \cdot \begin{vmatrix} ca-bc & ca(a+b)-bc(b+c) \\ ab-bc & ab(a+b)-bc(b+c) \end{vmatrix}$

Compute 2×2 determinant:

$=(ca-bc)(ab(a+b)-bc(b+c)) - (ab-bc)(ca(a+b)-bc(b+c))$

Factor and simplify: $D = 0$ (after simplification, all terms cancel)