Not sure if perpendicular is a clue

We can solve it by homogenizing the given circle with the given line. The equation of line can be written as :- $1 = \dfrac{x+2y}{3}$ Now we will replace $1$ with above expression so as to make the equation of circle homogeneous in $2^{nd}$ degree.

Hence the equation of circle becomes:- $x^{2} + y^{2} + (x-6y)\dfrac{(x+2y)}{3} + a\dfrac{(x+2y)}{9}^{2} =0$ Now, the above equation represents a pair of lines passing through origin and intersection point of the given line and circle.

Hence the two lines in above equation are $OP$ and $OQ$ . For $OP\perp OQ$ the sum of coefficients of $x^{2}$ and $y^{2}$ should be $0$ .

Coefficient of $x^{2} = 1+\dfrac{1}{3}+\dfrac{a}{9}$ .

Coefficient of $y^{2} = 1-4+\dfrac{4a}{9}$ .

Now their sum is $0$ hence $\boxed{a=3}$

Let the three points be $O(0,0)$ , $P(x_1,y_1)$ and $Q(x_2, y_2)$ .

$f(n) = \left\{ \begin{array}{l l} x+2y-3=0\\ x^2+y^2+x-6y+a=0 \end{array} \right.$

Substitute equation $(1)$ to $(2)$ ,

$\begin{aligned} (3-2y)^2+y^2+3-2y-6y+a&=0 \\ 5y^2-20y+12+a&=0\end{aligned}$

$\begin{aligned} \displaystyle y_1+y_2&=4 \\ \\ y_1y_2&=\frac{12+a}{5} \\ \end{aligned}$

Substitute $y=\frac{1}{2}(3-x)$ , we have

$\begin{aligned} x_1+x_2&=-2 \\ \\x_1x_2&=\frac{4a-27}{5} \end{aligned}$

Given that $OP\perp OQ$ , then

$\begin{aligned} \displaystyle m_{op} m_{oq} &=-1 \\ \\ \frac{y_1}{x_1} \frac{y_2}{x_2}&= -1 \\ \\ y_1 y_2 &= -x_1 x_2 \\ \\ \frac{12+a}{5} &= -\frac{4a-27}{5} \\ \\ a&=3 \end{aligned}$

First we solve for the coordinates of $P$ and $Q$ . From $x+2y=3\quad \Rightarrow x = 3 - 2y$ . Substitute this in the circle equation, we have:

$x^2+y^2+x-6y+a = (3-2y)^2+y^2+(3-2y) - 6y + a \\ \Rightarrow 5y^2-20y+12+a = 0\quad \Rightarrow y = 2 \pm \sqrt{1.6-0.2a} = 2 \pm \alpha$

$\Rightarrow y = \begin{cases} 2 + \alpha & \Rightarrow x = 3-2(2+\alpha) = -1-2 \alpha & \Rightarrow P(-1-2\alpha, 2+\alpha) \\ 2 - \alpha & \Rightarrow x = 3-2(2-\alpha) = -1+2 \alpha & \Rightarrow Q(-1+2\alpha, 2-\alpha) \end{cases}$

Let the gradient of $OP$ and $OQ$ be $m_p$ and $m_q$ respectively, and since $OP$ and $OQ$ are perpendicular, then we have:

$m_p = - \dfrac {1}{m_q} \quad \Rightarrow \dfrac {(2+\alpha)-0}{(-1-2\alpha)-0} = - \dfrac {(-1+\alpha)-0} {(2-\alpha)-0}$

$\Rightarrow (2+\alpha)(2-\alpha) = (1+2\alpha)(-1+\alpha) \quad \Rightarrow 4 - \alpha^2 = \alpha^2 -1 \\ \Rightarrow 5\alpha^2 = 5\quad \Rightarrow \alpha^2 = 1 \\ \Rightarrow 1.6 - 0.2a = 1 \quad \Rightarrow 0.2a = 0.6 \quad \Rightarrow a = \boxed{3}$

$Let~ C_1~be~given~\bigcirc~(x+\dfrac{1}{2})^2+(y-3)^2=9\frac{1}{4}-a=R^2,~~N~\odot~of~C_1.\\~~\\ C_2~be~\bigcirc~through~POQ. But \angle ~POQ=90^o\\\implies~PQ~is~its~diameter, M~the~\odot,~and~\therefore~MO=r,~radius~of~C_2. \\\color{#D61F06}{M~is~also~the~midpoint~of~chord~PQ~of~C_1.~~PM=r.} \\L_1~is ~line~y=-\dfrac{x}{2}+\dfrac{3}{2} ~ \therefore L_2 \perp ~L_1~through~N, is~y-3=2*(x+\dfrac{1}{2})\\\implies~L_2~is~y=2x+4.~~~~~~~~~~L_1~and~L_2~intersect~at~M,(-1,2)\\ \therefore NM^2=(-1-\dfrac{-1}{2})^2+(2-3)^2=\dfrac{5}{4}.~~r^2=MO^2=(-1)^2+2^2=5.\\~~\\\therefore~9\frac{1}{4}- a=R^2=NM^2+PM^2=NM^2+r^2=6\frac{1}{4}. ~~~~~~~a=~~~~~~~~~~\large \color{#3D99F6}{\boxed{3} }$