If $$a_r = \cos\frac{2r\pi}{9} + i\sin\frac{2r\pi}{9}$$, $$r = 1, 2, 3, \ldots$$, $$i = \sqrt{-1}$$, then the determinant $$\begin{vmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{vmatrix}$$ is equal to:

We begin by recognising that every number $$a_r$$ is written with the help of Euler’s formula. Set

$$\omega \;=\; \cos\dfrac{2\pi}{9} \;+\; i\sin\dfrac{2\pi}{9} \;=\;e^{\,i\;2\pi/9}.$$

Then, for every positive integer $$r$$, we have

$$a_r \;=\; \cos\dfrac{2r\pi}{9} \;+\; i\sin\dfrac{2r\pi}{9} \;=\;e^{\,i\;2r\pi/9} \;=\;\omega^{\,r}.$$

In particular, $$a_9=\omega^{\,9}=e^{\,i\,2\pi}=1$$ because $$e^{\,i\;2\pi}=1.$$ Substituting $$a_r=\omega^{\,r}$$ into the given determinant, we rewrite it purely in terms of powers of $$\omega$$:

$$\Delta \;=\; \begin{vmatrix} \omega^{\,1} & \omega^{\,2} & \omega^{\,3} \\ \omega^{\,4} & \omega^{\,5} & \omega^{\,6} \\ \omega^{\,7} & \omega^{\,8} & \omega^{\,9} \end{vmatrix} \;=\; \begin{vmatrix} \omega & \omega^{2} & \omega^{3} \\ \omega^{4} & \omega^{5} & \omega^{6} \\ \omega^{7} & \omega^{8} & 1 \end{vmatrix}.$$

Now we evaluate this $$3\times3$$ determinant by the cofactor expansion formula

$$ \begin{aligned} \Delta &=\; \omega \,\Bigl(\,\omega^{5}\cdot1-\omega^{6}\cdot\omega^{8}\Bigr) \;-\; \omega^{2} \,\Bigl(\,\omega^{4}\cdot1-\omega^{6}\cdot\omega^{7}\Bigr) \;+\; \omega^{3} \,\Bigl(\,\omega^{4}\cdot\omega^{8}-\omega^{5}\cdot\omega^{7}\Bigr). \end{aligned} $$

We simplify term by term, always using the index law $$\omega^{m}\,\omega^{n}=\omega^{\,m+n}$$ and the crucial fact $$\omega^{9}=1$$ (since raising $$\omega$$ to the ninth power brings the angle back to $$2\pi$$).

The first bracket evaluates to

$$\omega^{5}\cdot1-\omega^{6}\cdot\omega^{8} \;=\; \omega^{5}-\omega^{14} \;=\; \omega^{5}-\omega^{\,14-9} \;=\; \omega^{5}-\omega^{5} \;=\;0.$$

The second bracket becomes

$$\omega^{4}\cdot1-\omega^{6}\cdot\omega^{7} \;=\; \omega^{4}-\omega^{13} \;=\; \omega^{4}-\omega^{\,13-9} \;=\; \omega^{4}-\omega^{4} \;=\;0.$$

The third bracket gives

$$\omega^{4}\cdot\omega^{8}-\omega^{5}\cdot\omega^{7} \;=\; \omega^{12}-\omega^{12} \;=\;0.$$

Thus every cofactor expression inside the large parentheses is zero, and therefore

$$\Delta \;=\;\omega\,(0)\;-\;\omega^{2}\,(0)\;+\;\omega^{3}\,(0)\;=\;0.$$

The determinant itself vanishes, but among the four options offered, the expression that is always equal to the same value is

$$a_1a_9-a_3a_7 \;=\; \omega^{1}\cdot1-\omega^{3}\cdot\omega^{7} \;=\; \omega-\omega^{10} \;=\; \omega-\omega^{\,10-9} \;=\; \omega-\omega \;=\;0.$$

Hence the determinant equals this particular expression, corresponding to option B.

Hence, the correct answer is Option B.