Haitch Ke!

Since, they are collinear (lie on same line), the slope between any two points must be equal.

Slope of line joining $(3,3)$ and $(h,0)$ = Slope of line joining $(h,0)$ and $(0,k)$ .

---

$\dfrac{0-3}{h-3} = \dfrac{k-0}{0-h}$

$\dfrac{-3}{h-3} = \dfrac{k}{-h}$

$3h=k(h-3)$

$3h=kh-3k$

$3h+3k=kh$

Dividing both sides by $3kh$ , we get

$\dfrac{1}{h} + \dfrac{1}{k} = \boxed{\dfrac{1}{3}}$

One line passing through $(3,3)$ in $\mathbb{R}^2$ can be written as $y - 3 = m(x - 3) \Rightarrow \frac{3(m - 1)}{m} = h \space \text{ and } \space k = 3(1 - m) \Rightarrow$ $\frac{1}{h} + \frac{1}{k} = \frac{m}{3(m - 1)} - \frac{1}{3(m - 1)} = \boxed{\frac{1}{3}}$ Note.- If $m = 1$ , $h = k = 0$ , but, I'm supossing that they are 3 collinear points.