From the velocity-time graph, given in figure of a particle moving in a straight line, one can conclude that

$\text{ Displacement in }12\mathrm{s}=\text{ Area under }v-t\text{ graph }$

$=\frac{1}{2}\times (12+5)4=34\mathrm{m}$

$V_{\mathrm{av}}=\frac{\text{ Displacement }}{\text{ Time }}=\frac{34}{12}=\frac{17}{6}{\mathrm{ms}}^{-1}$

Hence, (a) is incorrect; (b) is incorrect because during first 3 s, velocity increases from 0 to 4 ms^-1 option; (c) is incorrect, because in part AB velocity is constant.