Delta = left| begin{array}{ccc} a & a+b & a+b | vedclass

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(A) Given $\Delta = \left| \begin{array}{ccc} a & a+b & a+b+c \ 3a & 4a+3b & 5a+4b+3c \ 6a & 9a+6b & 11a+9b+6c \end{array} ight|$.Applying row ope

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$i$

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(A) Given $\Delta = \left| \begin{array}{ccc} a & a+b & a+b+c \ 3a & 4a+3b & 5a+4b+3c \ 6a & 9a+6b & 11a+9b+6c \end{array} ight|$.Applying row operations $R_2 	o R_2 - 3R_1$ and $R_3 	o R_3 - 6R_1$:$\Delta = \left| \begin{array}{ccc} a & a+b & a+b+c \ 0 & a & 2a+b \ 0 & 3a & 5a+3b \end{array} ight|$.Expanding along the first column:$\Delta = a \left| \begin{array}{cc} a & 2a+b \ 3a & 5a+3b \end{array} ight| = a [a(5a+3b) - 3a(2a+b)]$.$\Delta = a [5a^2 + 3ab - 6a^2 - 3ab] = a[-a^2] = -a^3$.Given $a = i$,then $\Delta = -(i)^3 = -(-i) = i$.

$\Delta = \left| \begin{array}{ccc} a & a+b & a+b+c \\ 3a & 4a+3b & 5a+4b+3c \\ 6a & 9a+6b & 11a+9b+6c \end{array} \right|$ where $a = i, b = \omega, c = \omega^2$,then $\Delta$ is equal to

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