The system of linear equations $x + y + z = 2$, $2x + y - z = 3,$ $3x + 2y + kz = 4$has a unique solution if

Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ is equal to

Let $S_1$ and $S_2$ be respectively the sets of all $a \in R -\{0\}$ for which the system of linear equations

$a x+2 a y-3 a z=1$

$(2 a+1) x+(2 a+3) y+(a+1) z=2$

$(3 a+5) x+(a+5) y+(a+2) z=3$

has unique solution and infinitely many solutions. Then

The values of $x,y,z$ in order of the system of equations $3x + y + 2z = 3,$ $2x - 3y - z = - 3$, $x + 2y + z = 4,$ are

$\left| {\,\begin{array}{*{20}{c}}{1 + i}&{1 - i}&i\\{1 - i}&i&{1 + i}\\i&{1 + i}&{1 - i}\end{array}\,} \right| = $