From the figure

$l_{1}+l_{2}+l_{3}+l_{4}=$ constant

if block B moved by $X_{1}$ and block A moved by $X_{2}$ then $-$ moves down by $X_{4}$ and $l_{3}$ remain same)

Now $-\left(l_{1}-\left(X_{1}-X_{2}\right)\right)+l_{2}-\left(X_{1}-X_{2}\right)+l_{3}+\left(l_{4}+X_{4}\right)=\left(l_{1}+l_{2}+l_{3}+l_{4}\right)$

$\left(X_{1}+X_{2}\right)$ taken by the relative change in the position $\left(l_{1}+l_{2}-2\left(X_{1}-X_{2}\right)\right)+l_{3}+l_{4}+x_{4}=l_{1}+l_{2}+l_{3}+l_{4}$

$\rightarrow-2\left(X_{1}-X_{2}\right)=-X_{4}$

thus $X_{4}=2\left(X_{1}-X_{2}\right)$

derivate with respect to time

$\frac{d x_{4}}{d t}=2\left(\frac{d x_{1}}{d t}-\frac{d x_{2}}{d t}\right)$

$V_{4}+2\left(V_{1}-V_{2}\right)=2(3-1)=4 m / s$

because – block $'c'$ is also in th contact with $B.$

so the net velocity of the block $c$

$\vec{V}_{c}=-3 \hat{i}-4 \hat{j}$

speed $=\left|\vec{V}_{c}\right|=\sqrt{3^{2}+4^{2}}=5 m / s$