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 -
 

If ${\Delta _1} = \left| {\begin{array}{*{20}{c}}
  x&{\sin \,\theta }&{\cos \,\theta } \\ 
  {\sin \,\theta }&{ - x}&1 \\ 
  {\cos \,\theta }&1&x 
\end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}}
  x&{\sin \,2\theta }&{\cos \,\,2\theta } \\ 
  {\sin \,2\theta }&{ - x}&1 \\ 
  {\cos \,\,2\theta }&1&x 
\end{array}} \right|$, $x \ne 0$ ; then for all $\theta  \in \left( {0,\frac{\pi }{2}} \right)$

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

If $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha - b}\\b&c&{b\alpha - c}\\2&1&0\end{array}\,} \right| = 0$ and $\alpha \ne \frac{1}{2},$ then

For which of the following ordered pairs $(\mu, \delta)$ the system of linear equations  $x+2 y+3 z=1$ ; $3 x+4 y+5 z=\mu$ ; $4 x+4 y+4 z=\delta$ is inconsistent?

Let $\alpha, \beta$ and $\gamma$ be real numbers such that the system of linear equations

$x+2 y+3 z=\alpha$

$4 x+5 y+6 z=\beta$

$7 x+8 y+9 z=\gamma-$

is consistent. Let $| M |$ represent the determinant of the matrix

$M=\left[\begin{array}{ccc}\alpha & 2 & \gamma \\ \beta & 1 & 0 \\ -1 & 0 & 1\end{array}\right]$

Let $P$ be the plane containing all those $(\alpha, \beta, \gamma)$ for which the above system of linear equations is consistent, and $D$ be the square of the distance of the point $(0,1,0)$ from the plane $P$.

($1$) The value of $| M |$ is

($2$) The value of $D$ is