If $a,b,c$ are in harmonic progression, the straight line $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{1}{c}=0$ always passes through a fixed point, that point is
A. $(-1,-2)$
B. $(-1,2)$
C. $(1,-2)$
D. $(1,\dfrac{-1}{2})$

Hint: In this question, we are to find the fixed point through which the given straight line always passes. Since it is given that, the variables are in harmonic progression, we can use the harmonic mean condition such that the required point is extracted from the given equation.

Formula Used: The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.
If $a,b,c$ are in H.P, then the harmonic mean between $a$ and $c$ is given by $b=\dfrac{2ac}{a+c}$.
The equation of the line (intercept form) is $\dfrac{x}{a}+\dfrac{y}{b}=1$.

Complete step by step solution: Given that,
The values $a,b,c$ are in H.P.
So, we can write
$\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\text{ }...(1)$
Given that a straight line with the equation $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{1}{c}=0\text{ }...(2)$ passes through the required fixed point.
So, from (2),
\[\begin{align}
  & \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c} \\
 & \Rightarrow \dfrac{1}{c}=\dfrac{2}{b}-\dfrac{1}{a} \\
\end{align}\]
Then, substituting in (2), we get
$\begin{align}
  & \dfrac{x}{a}+\dfrac{y}{b}+\left( \dfrac{2}{b}-\dfrac{1}{a} \right)=0 \\
 & \Rightarrow \left( \dfrac{x}{a}-\dfrac{1}{a} \right)+\left( \dfrac{y}{b}+\dfrac{2}{b} \right)=0 \\
 & \Rightarrow \dfrac{1}{a}(x-1)+\dfrac{1}{b}(y+2)=0 \\
\end{align}$
Then, from this we can say that the straight always passes through a fixed point. I.e.,
$\begin{align}
  & x-1=0 \\
 & \Rightarrow x=1 \\
 & y+2=0 \\
 & \Rightarrow y=-2 \\
\end{align}$
Therefore, the required fixed point is $(1,-2)$.
We can also verify this fixed point by substituting in the given equation,
\[\begin{align}
  & \dfrac{x}{a}+\dfrac{y}{b}+\dfrac{1}{c}=0 \\
 & \Rightarrow \dfrac{1}{a}+\dfrac{-2}{b}+\dfrac{1}{c}=0 \\
 & \Rightarrow \dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2}{b} \\
\end{align}\]
Thus, the result is in H.P. Hence it is proved.
Consider $a,b,c$ as $\dfrac{1}{1},\dfrac{1}{2},\dfrac{1}{3}$. Then, on substituting in the given line, we get
$\begin{align}
  & \dfrac{x}{{}^{1}/{}_{1}}+\dfrac{y}{{}^{1}/{}_{2}}+\dfrac{1}{{}^{1}/{}_{3}}=0 \\
 & \Rightarrow x+2y+3=0 \\
\end{align}$
The obtained equation when passes through the obtained fixed point $(1,-2)$, we get the value of the equation as
$\begin{align}
  & =1+2(-2)+3 \\
 & =1-4+3 \\
 & =0 \\
\end{align}$
Thus, the required point is $(1,-2)$.

Option ‘C’ is correct

Note: Here we need to remember that the variables $a,b,c$ are in H.P. So, the condition for harmonic progression $\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}$ is used for finding the required point.