positions after motion ...
$A(2,3)$
$B(k+3,2k-1)$
$C(4k+4,13-k)$
for A, B, and C to be collinear, vectors AB and BC need to have the same direction.
slope for AB ... $\dfrac{2k-4}{1+k}$
slope for BC ... $\dfrac{14-3k}{3k+1}$
$\dfrac{2k-4}{1+k} = \dfrac{14-3k}{3k+1}$
solve the resulting quadratic for $k$