Using the properties of determinants, evaluate $\begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

Toolbox:
• If each element of a row(or column)of a determinant is multiplied by a constant 'k' then |A|=k|A|.
• If each element of a row(or a column)of A is expressed as the sum of two or more terms,then the determinants of A can be expressed as the sum of determinants of two or more matrices.
Let $\Delta=\begin{vmatrix}a-b-c & 2a &2a\\2b& b-c-a & 2b\\2c & 2c&c-a-b\end{vmatrix}$

Apply $R_1\rightarrow R_1+R_2+R_3$,we get

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

Let us take (a+b+c) as the common factor from $R_1$

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

Apply $C_2\rightarrow C_2-C_3$ and $C_3\rightarrow C_3-C_1$

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

On expanding along $R_1$ we get,

$\Delta=(a+b+c)[-(a+b+c)\times -(a+b+c)]$

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