left|begin{array}{ccc}a-b-c & 2a & 2a 2b & b | vedclass

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(D) Let $\Delta = \left|\begin{array}{ccc}a-b-c & 2a & 2a \ 2b & b-c-a & 2b \ 2c & 2c & c-a-b\end{array}ight|$.Applying $R_1 ightarrow R_1 + R_2

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$(a+b+c)^3$

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(D) Let $\Delta = \left|\begin{array}{ccc}a-b-c & 2a & 2a \ 2b & b-c-a & 2b \ 2c & 2c & c-a-b\end{array}ight|$.Applying $R_1 ightarrow R_1 + R_2 + R_3$,we get:$\Delta = \left|\begin{array}{ccc}a+b+c & a+b+c & a+b+c \ 2b & b-c-a & 2b \ 2c & 2c & c-a-b\end{array}ight|$.Taking $(a+b+c)$ common from $R_1$:$\Delta = (a+b+c) \left|\begin{array}{ccc}1 & 1 & 1 \ 2b & b-c-a & 2b \ 2c & 2c & c-a-b\end{array}ight|$.Applying $C_2 ightarrow C_2 - C_1$ and $C_3 ightarrow C_3 - C_1$:$\Delta = (a+b+c) \left|\begin{array}{ccc}1 & 0 & 0 \ 2b & -(a+b+c) & 0 \ 2c & 0 & -(a+b+c)\end{array}ight|$.Expanding along $R_1$:$\Delta = (a+b+c) [1 \cdot (-(a+b+c)) \cdot (-(a+b+c)) - 0]$$\Delta = (a+b+c) (a+b+c)^2 = (a+b+c)^3$.

$\left|\begin{array}{ccc}a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b\end{array}\right|$ is equal to

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