The distance between the line and the point is:
$$ d = \frac{ 30 \sqrt{ 314}}{ 157 } $$The distance from the point $ (x_0, y_0) $ to the line $ Ax + By + C = 0 $ is given by:
$$ d = \frac{ \left| Ax_0 + B_yo + C \right| }{ \sqrt{A^2 + B^2}} $$To apply this formula, we first need to express the line in standard form
$$\begin{aligned} y & = \frac{ 17 }{ 5 } x - 2 \\5 \cdot y &= 5 \cdot \left( \frac{ 17 }{ 5 } x - 2 \right) \\5 \cdot y &= 17 x - 10 \\17x-5y-10&=0\end{aligned}$$After substituting: $ A = 17 $ , $ B = -5 $ , $ C = -10 $ , $ x_0 = 5 $ and $ y_0 = 3 $ we have:
$$ \begin{aligned} d =& \frac{ \left| 17\cdot5 +\left(-5\right)\cdot3 + \left( -10\right) \right| }{\sqrt{ 17^2 + (-5)^2}} = \\ d =& \frac{ \left| 85 -15 -10 \right| }{\sqrt{ 289 + 25}} = \\ d =& \frac{ \left| 60 \right| }{\sqrt{ 314}} = \\ d =& \frac{ 60 }{ \sqrt{ 314 } } = \\ d =& \frac{ 30 \sqrt{ 314}}{ 157 } \end{aligned} $$