Evaluate $\begin{vmatrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix}$.

$(a + b + c)^3$

The correct answer is Option (3) → $(a + b + c)^3$ ##

We have,

$\begin{vmatrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix}$

$= \begin{vmatrix} a + b + c & a + b + c & a + b + c \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix} \quad [∵R_1 \to R_1 + R_2 + R_3]$

$= (a + b + c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix}$  [taking $(a + b + c)$ common from the first row]

$= (a + b + c) \begin{vmatrix} 0 & 0 & 1 \\ 0 & -(a + b + c) & 2b \\ (a + b + c) & (a + b + c) & (c - a - b) \end{vmatrix}$   $[∵C_1 \to C_1 - C_3 \text{ and } C_2 \to C_2 - C_3]$

On taking $(a + b + c)$ common from $C_1$ and $C_2$, we get

$= (a + b + c)^3 \begin{vmatrix} 0 & 0 & 1 \\ 0 & -1 & 2b \\ 1 & 1 & c - a - b \end{vmatrix}$

On expanding along $R_1$, we get

$= (a + b + c)^3 \cdot 1[0 - (-1)]$

$= (a + b + c)^3$