My approach:
Upon solving this Diophantine equation , I get my generalized solutions as:
$5(3k+7t) + 7(-2k-5t) = k$ where $t$ is an integer
Now its given that x and y are non-negative integers therefore
$3k+7t\geq0\therefore t\geq-\dfrac{3k}{7}$
$-2k-5t\geq0\therefore t\leq-\dfrac{2k}{5}$
therefore $-\dfrac{3k}{7}\leq t\leq-\dfrac{2k}{5}$ , giving a range of $\left(-\dfrac{2k}{5} + \dfrac{3k}{7}\right)+1 = \dfrac{k}{35}+1$ solutions.
Now how should I proceed further , how do I make cases on $k$ ?
It is not necessary to employ a lot of machinery to find the minimum value of $k$ (see below). The solutions $x,y\geq 0$ are described by
$$x=x_0+7t~,~ y=y_0-5t$$
for some $x_0,y_0$ that satisfy $5x_0+7y_0=k$. Suppose the seven solutions (that have to be contiguous in t) span the set $\{t_0,t_0+1,..., t_0+6\}$. Then it is evident that by demanding that all the (x,y) coordinates of the solutions be positive, that
$$x_0\geq -7t_0 ~~,~~ y_0\geq 5t_0+30$$
which implies that $k=5x_0+7y_0\geq 210$. We can easily check that the requirement is satisfied for $k=210$ with $x_0=0, y_0=30, t_0=0$, which shows that $k_\min=210$.