The number of values of k for which the system of linear equations...

For no solution

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

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

$⇒ {k}^{2} - 5k + 6 = 0 ⇒ k = 2, 3$

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

So, only $k = 3$ is possible.

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