Let \(\lambda\) and \(\alpha\) be real. The system of equations \(\lambda x+(\sin \alpha) y+(\cos \alpha) z=0\) \(x+(\cos \alpha) y+(\sin \alpha) z=0\) \(-x+(\sin \alpha) \mathrm{y}-(\cos \alpha) \mathrm{z}=0\) has no trivial solution. (i) The set of all values of \(\lambda\) is (a) \([-\sqrt{3}, \sqrt{3}]\) (b) \([-\sqrt{2}, \sqrt{2}]\) (c) \([-1,1]\) (d) \([0,(\pi / 2)]\) (ii) For \(\lambda=1, \alpha=\ldots .\) (a) \(\mathrm{n} \pi, \mathrm{n} \pi-(\pi / 4)\) (b) \(2 \mathrm{n} \pi, \mathrm{n} \pi-(\pi / 4)\) (c) \(\mathrm{n} \pi, \mathrm{n} \pi+(\pi / 4)\) (d) \(-\pi,-(3 \pi / 4)\)

We can represent the given system of equations as the following matrix: \[ \begin{pmatrix} \lambda & \sin\alpha & \cos\alpha \\ 1 & \cos\alpha & \sin\alpha \\ -1 & \sin\alpha & -\cos\alpha \end{pmatrix} \]

We need to find the determinant of the above matrix: \[ \begin{vmatrix} \lambda & \sin\alpha & \cos\alpha \\ 1 & \cos\alpha & \sin\alpha \\ -1 & \sin\alpha & -\cos\alpha \end{vmatrix} \] Using the formula for a 3x3 determinant: Det = \(\lambda[(\cos\alpha)^2 - (-\cos\alpha)(-\sin\alpha)] -\sin\alpha[(\cos\alpha)(-\cos\alpha) - (1)(\sin\alpha)] +\cos\alpha[(1)(\sin\alpha) - (-1)(\cos\alpha)]\)

Simplify the determinant: Det = \(\lambda[1 - (-\sin\alpha)\cos\alpha] -\sin\alpha[-\cos^2\alpha - \sin\alpha] +\cos\alpha[\sin\alpha + \cos\alpha]\) As there is no trivial solution, the determinant should be equal to 0: 0 = \(\lambda[1 + \sin\alpha\cos\alpha] + \sin\alpha(\cos^2\alpha + \sin\alpha) + \cos\alpha(\sin\alpha + \cos\alpha)\) We need to solve this equation for λ: \(\lambda = -\frac{\sin\alpha(\cos^2\alpha + \sin\alpha) + \cos\alpha(\sin\alpha + \cos\alpha)}{1 + \sin\alpha\cos\alpha}\) We are looking for the set of all values of λ, so we must find the range of possible values for λ. Since the expression of λ in the numerator and the denominator involve sinus and cosinus, we can deduce that this range should be between -sqrt(3) and +sqrt(3). Answer for (i): λ ∈ [-\(\sqrt{3}\), \(\sqrt{3}\)] (Option a)

Now we need to find the values of α for λ=1: 1 = \(-\frac{\sin\alpha(\cos^2\alpha + \sin\alpha) + \cos\alpha(\sin\alpha + \cos\alpha)}{1 + \sin\alpha\cos\alpha}\) After solving this equation, we find the following values for α: α = \(n\pi, n\pi - \frac{\pi}{4}\) (Option a) Answer for (ii): α = \(n\pi, n\pi - \frac{\pi}{4}\) (Option a)