We are given: \[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ bc & ca & ab \\ a(b+c) & b(c+a) & c(a+b) \end{vmatrix} \]

Expand the third row terms: \[ a(b+c) = ab + ac, \quad b(c+a) = bc + ab, \quad c(a+b) = ca + cb \] So the determinant becomes: \[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ bc & ca & ab \\ ab+ac & bc+ab & ca+bc \end{vmatrix} \]

Notice that the third row is the sum of the first two rows: \[ (ab+ac, \, bc+ab, \, ca+bc) = (ab,bc,ca) + (ac,ab,bc) \] which is simply a linear combination of Row 1 and Row 2.

Therefore, the rows of the determinant are linearly dependent. Hence: \[ \Delta = 0 \]

Final Answer: \( k = 0 \)
✅ Correct Option: (D)