Consider the parabola ${(x-1)}^2+{(y-2)}^2=\frac{{(12x-5y+3)}^2}{169}$

A) If parabola ${(x-1)}^2+{(y-2)}^2=\left|\frac{{(12x-5y+3)}^2}{169}\right|$

The locus of the point of intersection of perpendicular tangents is directrix which is  $12x-5y+3=0$

B) tangent at the vertex, which parallel to directrix and equidistant from the directrix and latusrectum  line

Let the equation of tangent at vertex be $12x-5y+\lambda =0$

$\Rightarrow \left|\frac{\lambda -3}{\sqrt{{12}^2+{(-5)}^2}}\right|=\left|\frac{\lambda +2}{\sqrt{{12}^2+5^2}}\right|$

$\Rightarrow \lambda -3=\pm (\lambda +2)$ 

$\Rightarrow \lambda -3=\lambda +2$    (does not exist)

$\Rightarrow \lambda -3=\lambda -2$

$\Rightarrow 2\lambda =1\Rightarrow \lambda =\frac{1}{2}$

Hence equation of tangent at vertex is  $24x-10y+1=0$.

C) The  minimum length of focal chord occurs along the latusrectum line, which is a line passing through the focus directrix i.e. $12x-5y-2=0$

D) The  parabola is symmetrical about its axis, which is a line passing through the focus $(1,2)$ and perpendicular to the directrix

So, it has equation $5x+12y-29=0$.