If non-zero numbers $$a, b, c$$ are in H.P., then the straight line $$\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$$ always passes through a fixed point. That point is

If $$a,\,b,\,c$$ are in H.P. (harmonic progression), then their reciprocals are in A.P. (arithmetic progression).
Thus, by definition of an A.P.,

$$\frac{2}{b}= \frac{1}{a} + \frac{1}{c}\,\,\,\,\,-(1)$$

The given straight-line equation is

$$\frac{x}{a} + \frac{y}{b} + \frac{1}{c}=0 \,\,\,\,\,-(2)$$

Let us test each option by substituting its coordinates $$\bigl(x_0,\,y_0\bigr)$$ into equation $$(2)$$ and then use relation $$(1)$$.

Case 1: Option A: $$(-1,\,2)$$
Substituting in $$(2):$$ $$\frac{-1}{a} + \frac{2}{b} + \frac{1}{c} = -\frac{1}{a} + \frac{2}{b} + \frac{1}{c}.$$
Using $$(1)$$, replace $$\frac{2}{b}$$ by $$\frac{1}{a}+\frac{1}{c}$$: $$-\frac{1}{a}+\Bigl(\frac{1}{a}+\frac{1}{c}\Bigr)+\frac{1}{c} =0 + \frac{2}{c}\neq 0.$$
Hence the line does not always pass through $$(-1,2).$$

Case 2: Option B: $$(-1,\,-2)$$
Substitute: $$\frac{-1}{a} - \frac{2}{b} + \frac{1}{c}.$$
Replace $$-\tfrac{2}{b}$$ with $$-\bigl(\tfrac{1}{a}+\tfrac{1}{c}\bigr):$$ $$-\frac{1}{a}-\Bigl(\frac{1}{a}+\frac{1}{c}\Bigr)+\frac{1}{c} = -\frac{2}{a}\neq 0.$$ So this point is also rejected.

Case 3: Option C: $$(1,\,-2)$$
Substitute: $$\frac{1}{a} - \frac{2}{b} + \frac{1}{c}.$$
Using $$(1):$$ $$-\frac{2}{b}= -\Bigl(\frac{1}{a}+\frac{1}{c}\Bigr).$$
Therefore $$\frac{1}{a} - \Bigl(\frac{1}{a}+\frac{1}{c}\Bigr) + \frac{1}{c} = \frac{1}{a}-\frac{1}{a}-\frac{1}{c}+\frac{1}{c}=0.$$ Hence the expression vanishes identically, so every such line passes through $$(1,-2).$$

Case 4: Option D: $$\left(1,\,-\tfrac{1}{2}\right)$$ can be checked similarly; it does not satisfy $$(2)$$ for arbitrary $$a,b,c$$ in H.P.

Therefore the fixed point common to all lines of the form $$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{1}{c}=0$$, under the given condition on $$a,b,c$$, is $$(1,-2).$$

Option C which is: $$(1,-2)$$