**Toolbox:**

- (i) If two rows or columns are identical then the value of the determinant is zero
- (ii)Elementary transformation can be made by
- (a) interchanging two rows or columns
- (b) By adding or subtracting two or more rows or columns.

Let $\Delta=\left|\begin{array}{ccc} 1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^{n} \\ \omega^n & \omega^{2n} & 1 \end{array}\right|$

We know $\omega^3=1\qquad Therefore\; \omega^{3n}=1$

Hence substitute for 1 in $R_1$ in the determinant we get

$\Delta=\left|\begin{array}{ccc} \omega^{3n} & \omega^{n} & \omega^{2n} \\ \omega^{2n} & 1 & \omega^{n} \\ \omega^n & \omega^{2n} & 1 \end{array}\right|$

Now let us take $\omega^n$ as the common factor from $R_1$

$\Delta=\omega^{n}\left|\begin{array}{ccc} \omega^{2n} & 1 & \omega^n \\ \omega^{2n} & 1 & \omega^{n} \\ \omega^n & \omega^{2n} & 1 \end{array}\right|$

Here two rows $R_1$ and $R_2$ are identical .

Hence the value of the determinant is zero

Therefore $\Delta=0$

Solution : option A is correct