Given curve $3 y=6 x-5 x^3$
$\Rightarrow \quad \frac{d y}{d x}=2-5 x^2$
Slope of normal $=\frac{-1}{2-5 x^2}$
Equation of normal with slope $\frac{-1}{2-5 x^2}$
passing through origin $(0,0)$.
$\frac{\left(y-y_1\right)}{x-x_1}=m$
$\Rightarrow \frac{y-0}{x-0}=\frac{-1}{2-5 x^2} \Rightarrow \frac{y}{x}=\frac{-1}{2-5 x^2}$
Since, $(h, k)$ lies on normal
$\frac{k}{h}=\frac{-1}{2-5 h^2}$
$\Rightarrow \frac{k}{h}=\frac{1}{5 h^2-2}$ ...(i)
Since, $(h, k)$ lies on curve
$3 k=6 h-5 h^3$
$\Rightarrow \quad 3 k=h\left(6-5 h^2\right)$
$\Rightarrow \quad \frac{k}{h}=\frac{6-5 h^2}{3}$ ...(ii)
On solving Eqs. (i) and (ii),
$\frac{1}{5 h^2-2}=\frac{6-5 h^2}{3}$
$\begin{aligned} & \Rightarrow \quad 3=\left(6-5 h^2\right)\left(5 h^2-2\right) \\ & \Rightarrow \quad 3=30 h^2-12-25 h^4+10 h^2 \\ & \Rightarrow \quad 15=-25 h^4+40 h^2 \\ & \Rightarrow \quad 5 h^4-8 h^2+3=0\end{aligned}$
For simplicity, let $h^2=z$
The equation becomes $5 z^2-8 z+3=0$
$(5 z-3)(z-1)=0$
Now, $z=\frac{3}{5}, 1$
Putting $h^2=z$
$\Rightarrow \quad h^2=\frac{3}{5} \Rightarrow h= \pm \sqrt{\frac{3}{5}}$
It is not in option.
or $(z-1)=0$
$z=1$
Putting $h^2=z$
$\Rightarrow \quad h^2=1$
$h= \pm 1$
But $h=-1$ is not in option.
$\therefore h=1$ is correct