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**Hint:**Using the section formula method we first multiply the \[x\] coordinates with the ratio \[3:2\] and then do the same for \[y\] coordinates as well.

For \[x\] coordinates and coordinates of A and B as \[\left( a,b \right)\] and \[\left( x,y \right)\] with ratio

\[\left( m:n \right)\]:

\[\dfrac{mx+na}{m+n}\]

For \[y\] coordinates and coordinates of A and B as \[\left( a,b \right)\] and \[\left( x,y \right)\] with ratio\[\left( m:n \right)\]:

\[\dfrac{my+nb}{m+n}\]

The previous formula was applied for P coordinate and now we will do the same for Q coordinates as well. The ratio of \[3:2\] will change to \[-3:2\] as Q is a harmonic conjugate of P.

**Complete step by step solution:**

Now as given in the question, we first form a coordinate diagram where P is the midpoint with A, B as extreme and the distance of AP is 3 and the distance of PB is 2.

After this let us form an equation with the help of A's coordinate and B's coordinate with midpoint as

\[\left( -1,4 \right)\].

The equation for the \[a\] or \[x\] coordinate is given as:

\[\Rightarrow \dfrac{3\times x+2\times a}{3+2}=-1\]

\[\Rightarrow 3x+2a=-5\] …(1)

The equation for the \[b\] or \[y\] coordinate is given as:

\[\Rightarrow \dfrac{3\times y+2\times b}{3+2}=4\]

\[\Rightarrow 3y+2b=20\] …(2)

We now find equation for \[x\] and \[y\] with midpoint being \[\left( 11,-8 \right)\] and the ratio of distance from AQ to QB as \[\left( -3:2 \right)\] as Q is harmonic conjugate of P.

The equation for the \[a\] or \[x\] coordinate is given as:

\[\Rightarrow \dfrac{-3\times x+2\times a}{-3+2}=11\]

\[\Rightarrow -3x+2a=-11\] …(3)

The equation for the \[b\] or \[y\] coordinate is given as:

\[\Rightarrow \dfrac{-3\times y+2\times b}{-3+2}=8\]

\[\Rightarrow -3y+2b=8\] …(4)

Now we equate the Equation 1, 2, 3 and 4; So as to find the value of a, b. First we find for the value of a by equating equation 1,3.

\[\begin{align}

& \text{ }3x+2a=-5 \\

& -3x+2a=-11 \\

& \text{ +}4a=-16 \\

\end{align}\]

\[\Rightarrow a=-4\]

We get the value of \[a\] as \[-4\] and to find the value of \[x\] we place the value of \[a\] in \[3x+2a= -5\].

\[\Rightarrow 3x+2\times -4=-5\]

\[\Rightarrow 3x=8-5\]

\[\Rightarrow x=1\]

Then we find for the value of b by Equating equation 2,4.

\[\begin{align}

& \text{ }3y+2b=20 \\

& -3y+2b=8 \\

& \text{ +}4b=28 \\

\end{align}\]

\[\Rightarrow b=7\]

We get the value of \[b\] as \[7\] and to find the value of \[x\] we place the value of \[b\] in \[-3y+2b=8\].

\[\Rightarrow -3y+2\times 7=8\]

\[\Rightarrow -3y=-14+8\]

\[\Rightarrow y=2\]

**Therefore, the value of a, b or A, B is given as \[\left( -4:7 \right)\]**

**Note:**The term harmonic conjugate means that if the line is divided let say in ratio of \[\text{AC:BC = AD:BD}\] we can say that C and D are cutting the line AB harmonically and that AB and CD are harmonic conjugates.

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