Let R = ( cases x & 0 & 0 0 & y & 0 0 & 0 & z cases ) be a non-zero...

Given,

$xsin\theta = ysin\left(\theta + \dfrac{2\pi}{3}\right) = zsin\left(\theta + \dfrac{4\pi}{3}\right) \ne 0$, 

$⇒ y = \dfrac{xsin\theta}{\sin \left(\theta + \dfrac{2\pi}{3}\right)}, z = \dfrac{xsin\theta}{\sin \left(\theta + \dfrac{4\pi}{3}\right)}$

Now, finding the trace of matrix we get,

$x + y + z = x + \dfrac{xsin\theta}{\sin \left(\theta + \dfrac{2\pi}{3}\right)} + \dfrac{xsin\theta}{\sin \left(\theta + \dfrac{4\pi}{3}\right)}$

$⇒ x + y + z = \dfrac{xsin\left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right) + xsin\theta sin\left(\theta + \dfrac{4\pi}{3}\right) + xsin\theta sin\left(\theta + \dfrac{2\pi}{3}\right)}{\sin \left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right)}$

$⇒ x + y + z = \dfrac{xsin\left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right) + xsin\theta\left(\sin \left(\theta + \dfrac{4\pi}{3}\right) + \sin \left(\theta + \dfrac{2\pi}{3}\right)\right)}{\sin \left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right)}$

$⇒ x + y + z = \dfrac{\dfrac{x}{2} \cdot 2\sin \left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right) + xsin\theta\left(2\sin \left(\theta + \pi\right)\cos \left(\dfrac{\pi}{3}\right)\right)}{\sin \left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right)}$

$⇒ x + y + z = \dfrac{\dfrac{x}{2} \cdot \left(\cos \left(\dfrac{2\pi}{3}\right) - \cos \left(2\theta + 2\pi\right)\right) + xsin\theta\left(- 2sin\theta \cdot \dfrac{1}{2}\right)}{\sin \left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right)}$

$⇒ x + y + z = \dfrac{\dfrac{x}{2} \cdot \left(\dfrac{- 1}{2} - \cos 2\theta\right) - x\sin^{2}\theta}{\sin \left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right)}$

$⇒ x + y + z = \dfrac{- x - 2xcos2\theta - 4x\sin^{2}\theta}{4\sin \left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right)}$

$⇒ x + y + z = \dfrac{- x - 2x\left(1 - 2\sin^{2}\theta\right) - 4x\sin^{2}\theta}{4\sin \left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right)}$

$⇒ x + y + z = \dfrac{- 3x + 4x\sin^{2}\theta - 4x\sin^{2}\theta}{4\sin \left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right)}$

$⇒ x + y + z = \dfrac{- 3x}{4\sin \left(\theta + \dfrac{2\pi}{3}\right)\sin \left(\theta + \dfrac{4\pi}{3}\right)} \ne 0$

\left(I\right)$Trace$( R ) = x + y + z \ne 0$</p><p>$⇒$Statement (i) is False</p><p>Now, finding$Adj ( Adj ( R ) )$</p><p>Now, using the property$Adj ( Adj ( R ) ) = R | R |$we get,</p><p>Trace ( Adj ( Adj ( R ) ) ) = xyz ( x + y + z ) \ne 0$

{As $x + y + z \ne 0$ proved above and $x, y \text{ and } z$ are non zero}

Hence, $Trace ( adj ( adj ( R ) ) = 0$ is a false statement,

So, neither \left(I\right)$nor \left(II\right)$ are true