If system of equations $kx + 2y – z = 2,$$\left( {k – 1} \right)x + ky + z = 1,x + \left( {k – 1} \right)y + kz = 3$ has only one solution, then number of possible real value$(s)$ of $k$ is –
 

The system of equations ${x_1} – {x_2} + {x_3} = 2,$ $\,3{x_1} – {x_2} + 2{x_3} = – 6$ and $3{x_1} + {x_2} + {x_3} = – 18$ has

$\left| {\,\begin{array}{*{20}{c}}{bc}&{bc' + b'c}&{b'c'}\\{ca}&{ca' + c'a}&{c'a'}\\{ab}&{ab' + a'b}&{a'b'}\end{array}\,} \right|$ is equal to

The number of distinct real roots of $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ in the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is

The values of $\alpha$, for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$, lie in the interval