Symmetry of point about a line in 3d How would I go about finding the symmetrical image of a 3d point $t = (t_x,t_y,t_z)$, about a 3d line given with the equation $\frac{x+1}{4}=\frac{y+1}{-3}=\frac{z-15}{16}$?

Edit: To clarify: The above illustration should show what I'm looking for. The coordinates of the point $t'$ on the graph. The purple line on the graph corresponds to the equation given above. It is a freshman linear algebra & analytic geometry exam question and a simple solution (without unneccesary differential equations or other complications) doable with pen and paper in a reasonable amount of time (of up to twenty minutes say) would be appreciated.

Let $\frac{x+1}{4}=\frac{y+1}{-3}=\frac{z-15}{16}=t$ and $(x,y,z)$ be the needed point.
Hence, $$\frac{x+t_x}{2}=-1+4t,$$
$$\frac{y+t_y}{2}=-1-3t$$ and
$$\frac{z+t_z}{2}=15+16t,$$ which gives,
$$x=-2+8t-t_x,$$
$$y=-2-6t-t_y$$ and
$$z=30+32t-t_z.$$
In another hand $$4(t_x-x)-3(t_y-y)+16(t_z-z)=0$$ and after substitution of values $x$, $y$ and $z$ in this equation we can find a value of $t$
and from here we can get values of $x$, $y$ and $z$.
I got $$t=\frac{4t_x-3t_y+16t_z-239}{281}$$
