Answer
$2i+4j+k.$
Work Step by Step
If a vector $v$ initiates at point $P_1(x_1,y_1,z_1)$ and terminates at $P_2(x_2,y_2,z_2)$ then $v$ is equal to the position vector $v=(x_2-x_1)i+(y_2-y_1)j+(z_2-z_1)k.$
Hence here $v=(5-3)i+(6-2)j+(0-(-1))k\\v=2i+4j+k.$