Given pair of perpendicular lines are given as,
$$
a x^2+6 x y+b y^2-10 x+10 y-6=0
$$

General equation is given as,
$$
a x^2+2 h x y+b y^2+2 g x+2 f y+c=0
$$

Compare Eqs. (i) and (ii), we obtain
$$
\begin{aligned}
& a=a, 2 h=6 \Rightarrow h=3, b=b, 2 g=-10 \Rightarrow g=-5 \\
& 2 f=10 \Rightarrow f=5, c=-6 \\
& \left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}
a & 3 & -5 \\
3 & b & 5 \\
-5 & 5 & -6
\end{array}\right|=0 \\
& \Rightarrow a(-6 b-25)-3(-18+25)-5(15-5 b)=0 \\
& \Rightarrow \quad 25 a+25 b+6 a b+96=0 \quad \ldots .
\end{aligned}
$$

Since, lines are perpendicular
$$
\Rightarrow \quad a+b=0 \text { or } a=-b
$$

Use Eq. (iv) in Eq. (iii),
$$
\begin{array}{rlrl}
& 25(a-a)+6 a(-a)+96 & =0 \\
\Rightarrow & & 6 a^2=96 \Rightarrow a^2 & =16 \\
\Rightarrow & & & a= \pm 4 \\
\Rightarrow & & & |a|=4
\end{array}
$$