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Let n be the number obtained on rolling a fair die. If the probability that the system $\mathrm{x}-\mathrm{ny}+\mathrm{z}=6 ; \mathrm{x}+(\mathrm{n}-2) \mathrm{y}+(\mathrm{n}+1) \mathrm{z}=8 ;(\mathrm{n}-1) \mathrm{y}+\mathrm{z}=1$ has a unique solution is $\frac{\mathrm{k}}{6}$, then the sum of $k$ and all possible values of n is
[JEE MAIN 22 Jan 2026 S II]
(1) 21
(2) 24
(3) 20
(4) 22Solution
$\mathrm{n} \in\{1,2,3,4,5,6\}$
$\Delta=\left|\begin{array}{ccc}1 & -n & 1 \\ 1 & n-2 & n+1 \\ 0 & n-1 & 1\end{array}\right|$
$\Delta=-(\mathrm{n}-1)(\mathrm{n}-2)$
For unique solution $\Delta \neq 0 \Rightarrow n \neq 1, n \neq 2$
favourable cases $n=\{3,4,5,6\}$
Required Probability $=\frac{4}{6}=\frac{\mathrm{k}}{6}$
$\mathrm{k}=4, \mathrm{n}=3,4,5,6$
required sum $=4+3+4+5+6=22$ Answer(4)
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