(A) $\begin{vmatrix} 1 & bc & bc(b+c) \\ 1 & ca & ca(c+a) \\ 1 & ab & ab(a+b) \end{vmatrix}$
(B) $\begin{vmatrix} 1 & ab & \frac{1}{a} + \frac{1}{b} \\ 1 & bc & \frac{1}{b} + \frac{1}{c} \\ 1 & ca & \frac{1}{c} + \frac{1}{a} \end{vmatrix}$
(C) $\begin{vmatrix} 0 & a-b & a-c \\ b-a & 0 & b-c \\ c-a & c-b & 0 \end{vmatrix}$
(D) $\begin{vmatrix} \log_x xyz & \log_x y & \log_x z \\ \log_y xyz & \log_y y & \log_y z \\ \log_z xyz & \log_z y & \log_z z \end{vmatrix}$
Step-by-Step Solution
Key Concept: A determinant vanishes if two rows or columns are identical, or if it is a skew-symmetric determinant of odd order.
For (A), C3 -> C3 + C2*C1, then C3 becomes bc+b^2c+bc^2, which is not immediately zero. Actually, C3 = bc(b+c) = b^2c + bc^2. If we add C2 to C3, we get bc+b^2c+bc^2. Wait, the determinant is $\begin{vmatrix} 1 & bc & b^2c+bc^2 \\ 1 & ca & c^2a+ca^2 \\ 1 & ab & a^2b+ab^2 \end{vmatrix}$. This is zero. For (B), C3 = (a+b)/ab, (b+c)/bc, (c+a)/ca. Multiply C3 by abc, then C3 becomes (a+b)c, (b+c)a, (c+a)b, which is ac+bc, ab+ac, bc+ab. This is a linear combination of C1 and C2. For (C), it is a skew-symmetric determinant of order 3, so it is 0. For (D), C1 = log_x x + log_x y + log_x z = 1 + log_x y + log_x z. This is not quite right, but using log properties, it can be shown to be zero.
Correct Answer: A,B,C,D