Let omega = - frac{1}{2} + ifrac{sqrt{3}}{2}. | vedclass

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(B) Given $\omega = - \frac{1}{2} + i\frac{\sqrt{3}}{2}$,we know that $\omega$ is a complex cube root of unity,so $1 + \omega + \omega^2 = 0$ and $\om

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$3\omega(\omega - 1)$

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(B) Given $\omega = - \frac{1}{2} + i\frac{\sqrt{3}}{2}$,we know that $\omega$ is a complex cube root of unity,so $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. Also,$\omega^4 = \omega^3 \cdot \omega = \omega$.Let $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & -1 - \omega^2 & \omega^2 \ 1 & \omega^2 & \omega^4 \end{array} ight|$.Using the property $1 + \omega + \omega^2 = 0$,we have $-1 - \omega^2 = \omega$.Substituting this and $\omega^4 = \omega$ into the determinant:$\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{array} ight|$.Applying the column operation $C_1 	o C_1 + C_2 + C_3$:$\Delta = \left| \begin{array}{ccc} 1+1+1 & 1 & 1 \ 1+\omega+\omega^2 & \omega & \omega^2 \ 1+\omega^2+\omega & \omega^2 & \omega \end{array} ight| = \left| \begin{array}{ccc} 3 & 1 & 1 \ 0 & \omega & \omega^2 \ 0 & \omega^2 & \omega \end{array} ight|$.Expanding along the first column:$\Delta = 3(\omega \cdot \omega - \omega^2 \cdot \omega^2) = 3(\omega^2 - \omega^4) = 3(\omega^2 - \omega)$.Factoring out $-\omega$:$\Delta = 3\omega(\omega - 1)$.

List $I$	List $II$
$P.$ Let $y(x)=\cos \left(3 \cos ^{-1} x\right), x \in[-1,1], x \neq \pm \frac{\sqrt{3}}{2}$. Then $\frac{1}{y(x)}\left\{\left(x^2-1\right) \frac{d^2 y(x)}{d x^2}+x \frac{d y(x)}{d x}\right\}$ equals	$1. \ 1$
$Q.$ Let $A_1, A_2, \ldots, A_n(n>2)$ be the vertices of a regular polygon of $n$ sides with its centre at the origin. Let $\vec{a}_k$ be the position vector of the point $A_k, k=1,2, \ldots, n$. If $\left\|\sum_{k=1}^{n-1}\left(\vec{a}_k \times \vec{a}_{k+1}\right)\right\|=\left\|\sum_{k=1}^{n-1}\left(\vec{a}_k \cdot \vec{a}_{k+1}\right)\right\|$,then the minimum value of $n$ is	$2. \ 2$
$R.$ If the normal from the point $P(h, 1)$ on the ellipse $\frac{x^2}{6}+\frac{y^2}{3}=1$ is perpendicular to the line $x+y=8$,then the value of $h$ is	$3. \ 8$
$S.$ Number of positive solutions satisfying the equation $\tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right)$ is	$4. \ 9$

Let $\omega = - \frac{1}{2} + i\frac{\sqrt{3}}{2}$. Then the value of the determinant $\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 - \omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4 \end{array} \right|$ is

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