Let $\mathrm{A}=\left(\begin{array}{ccc}{[\mathrm{x}+1]} & {[\mathrm{x}+2]} & {[\mathrm{x}+3]} \\ {[\mathrm{x}]} & {[\mathrm{x}+3]} & {[\mathrm{x}+3]} \\ {[\mathrm{x}]} & {[\mathrm{x}+2]} & {[\mathrm{x}+4]}\end{array}\right)$, where $[\mathrm{t}]$ 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:
Correct Option: , 2
$\left|\begin{array}{ccc}{[x+1]} & {[x+2]} & {[x+3]} \\ {[x]} & {[x+3]} & {[x+3]} \\ {[x]} & {[x+2]} & {[x+4]}\end{array}\right|=192$
$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3} \& \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$
$\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & -1 \\ {[\mathrm{x}]} & {[\mathrm{x}]+2} & {[\mathrm{x}]+4}\end{array}\right]=192$
$2[x]+6+[x]=192 \Rightarrow[x]=62$
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