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The origin remains at the same point, so the equation of axis of rotation can be written as $\frac{x-a}{a}=\frac{y-b}{b}=\frac{z-c}{c}$ It can be observed that the point obtained after rotation will lie on the plane passing through the original point with axis of rotation as normal to it.

$(4,-1,8)$ will lie on $(x-6)(a)+(y-6)(b)+(z-3)(c)=0$ and $(4,8,-1)$ will lie on $(x+6)(a)+(y-3)(b)+(z-6)(c)=0$ .

On substituting the values, we obtain

$2a+7b-5c=0\\10a+5b-7c=0$

On eliminating $a$ from both the equations, we get $b=\frac{3}{5}c$

$\begin{aligned} \frac{b+c}{a}&=\frac{b+c}{\frac{5c-7b}{2}}\\&=\frac{\frac{3}{5}c+c}{\frac{5c-\frac{21}{5}c}{2}}\\&=2\frac{\frac{8}{5}}{\frac{4}{5}}\\&\huge\color{#3D99F6}{=\boxed{4}}\end{aligned}$

As @Rohit Ner observes, for any point $P$ in space, the vector from $P$ to $T(P)$ is perpendicular to the axis $L$ of rotation. For the two given points, these vectors are $(4,-1,8)-(6,6,3)$ $=(-2,-7,5)$ and $(4,8,-1)-(-6,3,6)=(10,5,-7)$ . The vector product $(-2,-7,5) \times (10,5,-7) = (24,36,60)$ is parallel to $L$ . Thus $L$ consists of the points $(a,b,c)=(24t,36t,60t)$ and $\frac{b+c}{a}=\frac{96t}{24t}=\boxed{4}$