Solution

Comparing $$ 5x+2y=2k$$ with $$a_1x+b_1y+c_1=0$$, we get,
$$a_1=5, b_1=2,c_1=-2k$$

Comparing $$2(k+1)x+ky=3k+4$$ with $$a_2x+b_2y+c_2=0$$, we get,
$$a_2=2(k+1), b_2=k, c_2=-(3k+4)$$

Now,
Given: the equations has infinite no of solutions.
$$\therefore \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$

$$\dfrac{5}{2(k+1)}=\dfrac{2}{k}=\dfrac{-2k}{-(3k+4)}$$

When,

$$\dfrac{5}{2(k+1)}=\dfrac{2}{k}$$

$$\implies 5k=4(k+1)$$
$$\implies 5k-4k=4$$
$$\implies k=4$$

and

$$\dfrac{2}{k}=\dfrac{-2k}{-(3k+4)}$$

$$\implies 2(3k+4)=2k^2$$
$$\implies 3k+4=k^2$$
$$\implies k^2-3k-4=0$$
$$\implies k^2-4k+k-4=0$$
$$\implies k(k-4)+(k-4)=0$$
$$\implies (k-4)(k+1)=0$$

Either, $$k=4\space or\space k=-1$$
$$k=4$$ is the value that satisfies both the conditions.

So, $$Option \ A \ is\ correct$$