The first step in determining the regression line is to calculate the mean values of $X$ and $Y$.
$$\begin{array}{rcl} \bar{X}&=&\cfrac{\sum\limits_{i=1}^n{X_i}}{n} = \dfrac{1+2+3+4+5}{5}=\dfrac{15}{5}=3\\\\ \bar{Y}&=&\cfrac{\sum\limits_{i=1}^n{Y_i}}{n} = \dfrac{3+1+2+4+0}{5}=\dfrac{10}{5}=2 \end{array}$$
Next, find the values of $(X_i-\bar{X}), (Y_i-\bar{Y}), (X_i-\bar{X})^2$ and $(X_i-\bar{X})(Y_i-\bar{Y})$ for each pair of data points:
$$\begin{array}{|c|c|c|c|c|c|} \hline X_i&Y_i&X_i-\bar{X}&Y_i-\bar{Y}&(X_i-\bar{X})^2&(X_i-\bar{X})(Y_i-\bar{Y})\\ \hline 1&3&-2&1&4&-2\\ 2&1&-1&-1&1&1\\ 3&2&0&0&0&0\\ 4&4&1&2&1&2\\ 5&0&2&-2&4&-4\\ \hline \end{array}$$
With this information the slope $b_1$ and intercept $b_0$ can be calculated:
$$\begin{array}{rcl} b_1 &=& \cfrac{\sum\limits_{i=1}^n{(X_i-\bar{X})(Y_i-\bar{Y})}}{\sum\limits_{i=1}^n{(X_i-\bar{X})^2}} = \cfrac{-2+1+0+2-4}{4+1+0+1+4} = -0.3\\\\ b_0 &=& \bar{Y} - b_1 \cdot \bar{X} = 2 - (-0.3)\cdot3 = 2.9 \end{array}$$
So the regression equation is:
$$\hat{Y}=b_0 + b_1X=2.9 -0.3X$$
Note that the regression line always passes through the point $(\bar{X},\bar{Y})$. In this case, $(\bar{X},\bar{Y})= (3,2)$ and entering $X=3$ into the equation gives:
$$\hat{Y}=2.9 -0.3\cdot 3 =2$$