Let $S$ be the set of all $\lambda \in \mathrm{R}$ for which the system of linear equations

$2 x-y+2 z=2$

$x-2 y+\lambda z=-4$

$x+\lambda y+z=4$

has no solution. Then the set $S$

Let $a, b, c > 0$ and $\Delta  = \left| \begin{gathered}
  a + b\,\,b\,\,c \hfill \\
  b\, + \,c\,\,c\,\,\,a \hfill \\
  c + a\,\,a\,\,b \hfill \\ 
\end{gathered}  \right| ,$ then which of the following is not correct?

Let $M$ and $N$ be two $3 \times 3$ matrices such that $M N=N M$. Further, if $M \neq N^2$ and $M^2=N^4$, then

$(A)$ determinant of $\left( M ^2+ MN ^2\right)$ is $0$

$(B)$ there is a $3 \times 3$ non-zero matrix $U$ such that $\left( M ^2+ MN ^2\right) U$ is the zero matrix

$(C)$ determinant of $\left( M ^2+ MN ^2\right) \geq 1$

$(D)$ for a $3 \times 3$ matrix $U$, if $\left( M ^2+ MN ^2\right) U$ equals the zero matrix then $U$ is the zero matrix

If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&3&5\\2&{x + 2}&5\\2&3&{x + 4}\end{array}\,} \right| = 0$, then $ x =$

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

Let $P $ and $Q $ be $3×3$ matrices $P \ne Q$. If ${P^3} = {Q^3},{P^2}Q = {Q^2}P$ then determinant of $\det \left( {{P^2} + {Q^2}} \right)$ is equal to :