Let $A=\left(\begin{array}{ccc}{[x+1]} & {[x+2]} & {[x+3]} \\ {[x]} & {[x+3]} & {[x+3]} \\ {[x]} & {[x+2]} & {[x+4]}\end{array}\right),$ where $[t]$ denotes the greatest integer less than or equal to $\mathrm{t}$. If $\operatorname{det}(\mathrm{A})=192$, then the set of values of $\mathrm{x}$ is the interval

If ${A_i} = \left[ {\begin{array}{*{20}{c}}{{a^i}}&{{b^i}}\\{{b^i}}&{{a^i}}\end{array}} \right]$and if $|a|\, < 1,\,|b|\, < 1$, then $\sum\limits_{i = 1}^\infty {\det ({A_i})} $is equal to

$\left| {\,\begin{array}{*{20}{c}}{{{\sin }^2}x}&{{{\cos }^2}x}&1\\{{{\cos }^2}x}&{{{\sin }^2}x}&1\\{ - 10}&{12}&2\end{array}\,} \right| = $

if $\left| \begin{gathered}
   - 6\ \ \,\,1\ \ \,\,\lambda \ \  \hfill \\
  \,0\ \ \,\,\,\,3\ \ \,\,7\ \  \hfill \\
   - 1\ \ \,\,0\ \ \,\,5\ \  \hfill \\ 
\end{gathered}  \right| = 5948 $, then $\lambda $  is

If the system of linear equations $x + ky + 3z = 0;3x + ky - 2z = 0$ ; $2x + 4y - 3z = 0$  has a non-zero solution $\left( {x,y,z} \right)$ then $\frac{{xz}}{{{y^2}}} = $. . . . .

If the system of equations

$ 2 x+7 y+\lambda z=3 $

$ 3 x+2 y+5 z=4 $

$ x+\mu y+32 z=-1$

has infinitely many solutions, then $(\lambda-\mu)$ is equal to $\qquad$