Step 1: For infinitely many solutions, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$.
Step 2: Here $a_1=k,\ b_1=-1,\ c_1=-2$ and $a_2=6,\ b_2=-2,\ c_2=-3$.
Step 3: From $\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ we get $\dfrac{-1}{-2}=\dfrac{-2}{-3}$, i.e. $\dfrac{1}{2}=\dfrac{2}{3}$, which is false.
Step 4: Since the last two ratios can never be equal, no value of $k$ can make all three ratios equal.
Answer: (D) Such a value of $k$ does not exist.