Mathematics - Determinant Question with Solution | TestHub

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$\frac{k+2}{k}=\frac{10}{k+3}=\frac{k}{k-1}$

$\Rightarrow \left(k+2\right)\left(k+3\right)=10k$

$\Rightarrow k^2-5k+6=0\Rightarrow k=2, 3$

$k\ne 2$ for $k=2$ both lines are identical.

So, only $k=3$ is possible.

So the number of values of $k$ is $1$.