From the RAG we can make the necessary matrices.

$$\overset{\text{Allocation}}{\begin{array}{|l|l|l|l|}\hline \text{} & r_1 &r_2 & r_3 \\ \hline P_0 & \text{1} & \text{0} & \text{1} \\\hline P_1 & \text{1} & \text{1} & \text{0} \\\hline P_2 & \text{0} & \text{1} & \text{0} \\\hline P_3& \text{0} & \text{1} & \text{0} \\\hline \end{array}}\qquad \overset{\text{Future Need}}{\begin{array}{|l|l|l|l|}\hline \text{} &r_1 & r_2 & r_3 \\ \hline P_0 & \text{0} & \text{1} & \text{1} \\\hline P_1 & \text{1} & \text{0} & \text{0} \\\hline P_2 & \text{0} & \text{0} & \text{1} \\\hline P_3 & \text{1} & \text{2} & \text{0} \\\hline \end{array}}$$

- $\text{Total}=(2\quad 3\quad 2)$
- $\text{Allocated}=(2\quad 3\quad 1)$
- $\text{Available}=\text{Total} -\text{Allocated} =(0\quad 0\quad 1)$

$P_2's$ need $(0\quad 0\quad 1 )$ can be met

And it releases its held resources after running to completion

$A=(0\quad 0\quad 1)+(0\quad 1\quad 0)=(0\quad 1\quad 1)$

$P_0's$ need $(0\quad 1\quad 1 )$ can be met

and it releases

$A=(0\quad 1\quad 1)+(1\quad 0\quad 1)=(1\quad 1\quad 2)$

$P_1's$ needs can be met $(1\quad 0\quad 0)$ and it releases

$A=(1\quad 1\quad 2)+(1\quad 1\quad 0)=(2\quad 2\quad 2)$

$P_3's$ need can be met

So, the safe sequence will be $P_2-P_0- P_1- P_3.$