Let the directrix of the parabola $$P: y^2 = 8x$$ cuts the x-axis at the point $$A$$.Let $$B(\alpha, \beta)$$, $$\alpha > 1$$, be a point on $$P$$ such that the  slope of $$AB$$ is $$3/5$$. If  $$BC$$ is a focal chord of chord of $$P$$. then six times the area off $$(\triangle ABC)$$ is :

The given parabola is $$y^{2}=8x$$.
Write it in the standard form $$y^{2}=4ax$$ to get $$4a=8 \;\Rightarrow\; a=2$$.

For a right-opening parabola:
Focus $$S(a,0)=(2,0)$$, and directrix $$x=-a=-2$$.

The directrix meets the $$x$$-axis at
$$A(-2,0).$$

Let the required point on the parabola be $$B(\alpha,\beta)$$ with $$\alpha\gt 1$$.
Slope of $$AB$$ is given to be $$\dfrac{3}{5}$$, so $$\text{slope}(AB)=\frac{\beta-0}{\alpha-(-2)}=\frac{\beta}{\alpha+2}=\frac{3}{5}$$ $$\Longrightarrow\; \beta=\frac{3}{5}(\alpha+2).$$

Because $$B$$ lies on $$y^{2}=8x$$, substitute $$\beta$$: $$\left[\frac{3}{5}(\alpha+2)\right]^{2}=8\alpha$$ $$\frac{9(\alpha+2)^{2}}{25}=8\alpha$$ $$9(\alpha+2)^{2}=200\alpha.$$

Expand and rearrange: $$9\alpha^{2}+36\alpha+36-200\alpha=0$$ $$9\alpha^{2}-164\alpha+36=0.$$

Solve the quadratic: Discriminant $$\Delta=(-164)^{2}-4\cdot9\cdot36=25600 \; \Rightarrow\; \sqrt{\Delta}=160$$ $$\alpha=\frac{164\pm160}{18} \; \Rightarrow\; \alpha_{1}=18,\; \alpha_{2}=\frac{2}{9}.$$ Since $$\alpha\gt1$$, choose $$\alpha=18$$.
Then $$\beta=\frac{3}{5}(18+2)=12$$, so $$B(18,12).$$

The chord $$BC$$ is a focal chord, i.e., it passes through the focus $$S(2,0)$$.
Slope $$m_{SB}=\dfrac{12-0}{18-2}=\dfrac{3}{4}.$br/> Equation of $$SB$$ (and hence of $$BC$$): $$y=$$\frac{3}{4}$$(x-2) \; \Longrightarrow\; y=$$\frac{3}{4}$$x-$$\frac{3}{2}$$.$$

Find the second intersection $$C(x,y)$$ of this line with the parabola:

Substitute $$y$$ in $$y^{2}=8x$$: $$$$\left$$($$\frac{3}{4}$$x-$$\frac{3}{2}$$$$\right$$)^{2}=8x$$ $$$$\frac{9(x-2)^{2}$$}{16}=8x$$ $$9(x-2)^{2}=128x$$ $$9x^{2}-164x+36=0.$$

This quadratic already has root $$x=18$$ (point $$B$$); the other root is $$x_C=$$\frac{2}{9}$$.$$ Coordinates of $$C$$: $$y_C=$$\frac{3}{4}$$\!$$\left$$($$\frac{2}{9}$$$$\right$$)-$$\frac{3}{2}= \frac{1}{6}-\frac{9}{6}$$=-$$\frac{4}{3}$$.$$ Thus $$C\!$$\left$$($$\frac{2}{9}$$,-$$\frac{4}{3}$$$$\right$$).$$

Now compute the area of $$\triangle ABC$$ with vertices $$A(-2,0),\; B(18,12),\; C\!$$\left$$($$\frac{2}{9}$$,-$$\frac{4}{3}$$$$\right$$).$$

Using the determinant (shoelace) formula: $$$$\text{Area}=\frac$$12$$\left$$|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$$\right$$|$$ $$=$$\frac$$12$$\left$$[(-2)\!$$\left$$(12+$$\frac{4}{3}$$$$\right$$)+18\!$$\left$$(-$$\frac{4}{3}$$-0$$\right$$)+$$\frac{2}{9}$$(0-12)$$\right$$]$$ $$=$$\frac$$12$$\left$$[(-2)\!$$\left$$($$\frac{40}{3}$$$$\right$$)+18\!$$\left$$(-$$\frac{4}{3}$$$$\right$$)-$$\frac{24}{9}$$$$\right$$]$$ $$=$$\frac$$12$$\left$$[\,-$$\frac{80}{3}$$-24-$$\frac{8}{3}$$$$\right$$]$$ $$=$$\frac$$12$$\left$$(-$$\frac{160}{3}$$$$\right$$)=$$\frac{80}{3}$$.$$

Hence $$6$$\times$$$$\text{Area}$$=6$$\left$$($$\frac{80}{3}$$$$\right$$)=160.$$

Therefore, $$6 $$\times$$ $$\text{area}$$(\triangle ABC)=160,$$ which corresponds to
Option B : 160.