Given parabola is $\mathrm{y}^2=8 \mathrm{x}$ with parametric coordinates $\left(\mathrm{at}_1^2, 2 \mathrm{at}{ }_1\right)$ and $\left(\mathrm{at}_2^2, 2 \mathrm{at}_2\right)$.
$\operatorname{Let}\left(\mathrm{x}_1, \mathrm{y}_1\right) \rightarrow\left(a \mathrm{t}_1^2, 2 a t_1\right)$ and $\left(\mathrm{x}_2, \mathrm{y}_2\right) \rightarrow\left(a t_2^2, 2 a t_2\right)$
Equation of line passing through
$$
\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { and }\left(\mathrm{x}_2, \mathrm{y}_2\right) \text { is }\left(\mathrm{y}-2 \mathrm{at}_1\right)
$$


$\begin{aligned} & =\frac{2 a t_2-2 a t_1}{a t_2^2-a t_1^2}\left(x-a t_1^2\right) \\ & \left(y-2 a t_1\right)=\frac{2}{\left(t_2+t_1\right)}\left(x-a t_1^2\right)\end{aligned}$
Satisfy $(1,2)$,
$$
\begin{aligned}
& \left(2-2 a t_1\right)=\frac{2}{\left(t_2+t_1\right)}\left(1-a t_1^2\right) \\
& \left(1-a_1\right)=\frac{1}{\left(t_2+t_1\right)}\left(1-a_1^2\right) \\
& \left(1-a t_1\right)\left(t_2+t_1\right)=\left(1-a t_1^2\right) \\
& \mathrm{t}_2+t_1-\mathrm{at}_1 \mathrm{t}_2-\mathrm{at}_1^2=1-\mathrm{at}_1^2
\end{aligned}
$$
Here foci $a=2$.

When satisfy foci coordinates $(a, 0)$ then the required equation gives the relation $t_1 t_2=-1$
From (i)
$$
\begin{aligned}
& \mathrm{t}_2+\mathrm{t}_1-2(-1)=1 \\
& \mathrm{t}_2+\mathrm{t}_1+2=1
\end{aligned}
$$

$$
\begin{aligned}
& \mathrm{x}_1+\mathrm{x}_2=a \mathrm{t}_1^2+a \mathrm{t}_2^2 \\
& =\mathrm{a}\left[\mathrm{t}_1^2+\mathrm{t}_2^2+2 \mathrm{t}_1 \mathrm{t}_2-2 \mathrm{t}_1 \mathrm{t}_2\right] \\
& =\mathrm{a}\left[\left(\mathrm{t}_1+\mathrm{t}_2\right)^2-2 \mathrm{t}_1 \mathrm{t}_2\right] \\
& =2\left[(-1)^2-2(-1)\right] \\
& =[1+2]=2 \times 3=6
\end{aligned}
$$
So, option (c) is correct.