Given circle is $x^2+y^2-6x-2y+1=0$.

The tangent at $\left(a,4\right)$ is

$S\equiv \left(x\right)\left(a\right)+y\left(4\right)-3(x+a)-(y+4)+1=0$

$\Rightarrow xa+4y-3x-3a-y-4+1=0$

$\Rightarrow \left(a-3\right)x+3y-3a-3=0$

$\Rightarrow \frac{a-3}{3}x+y-a-1=0$

The tangent is $y+c=0$.

Comparing the coefficients we get,

$\frac{a-3}{3}=0\Rightarrow a-3=0\Rightarrow a=3$

$\Rightarrow -a-1=c\Rightarrow c=-3-1=-4$

$\Rightarrow ac=3(-4)=-12$