Question

If $\begin{vmatrix}a-b-c&2a&2a\\ 2b&b-c-a&2b\\ 2c&2c&c-a-b\end{vmatrix} = (a + b + c) (x + a + b + c)^2, x \ne 0$ and $a + b + c \ne 0$, then $x$ is equal to :

• A. $-(a + b + c)$
• B. $2(a + b + c)$
• C. $abc$
• D. $-2(a + b + c)$

Answer 1

Answer: (D)

$\begin{vmatrix}a-b-c&2a&2a\\ 2b&b-c-a&2b\\ 2c&2c&c-a-b\end{vmatrix}$
$ R_{1} \to R_{1} + R_{2} +R_{3} $
$ = \begin{vmatrix}a+b+c&a+b+c&a+b+c\\ 2b&b-c-a&2b\\ 2c&2c&c-a-b\end{vmatrix} $
$=\left(a+b+c\right) \begin{vmatrix}1&0&0\\ 2b&-\left(a+b+c\right)&0\\ 2c&2c&c-a-b\end{vmatrix} $
$= \left(a + b + c\right)\left(a + b + c\right)^{2 } $
$ \Rightarrow x = -2\left(a + b + c\right)$