Answer
$(-7,-2,-7); (\dfrac{19}{3}, \dfrac{14}{3},\dfrac{19}{3})$
Work Step by Step
Formula to calculate normal line equation is:
$\dfrac{(x_2-x_1)}{f_x(x_1,y_1,z_1)}=\dfrac{(y_2-y_1)}{f_y(x_1,y_1,z_1)}=\dfrac{(z_2-z_1)}{f_x(x_1,y_1,z_1)}$
At point$(1,2,1)$
$\dfrac{(x-1)}{8}=\dfrac{(y-2)}{4}=\dfrac{(z-1)}{8}$
Consider
$\dfrac{(x-1)}{8}=\dfrac{(y-2)}{4}=\dfrac{(z-1)}{8}=k$
Therefore, $x=1+8k;y=2+4k,z=1+8k$
Plug these values into the equation of the sphere: : $x^2+y^2+z^2=102$
After simplifications we get $t=-1,\dfrac{2}{3}$
The desired points are: $(-7,-2,-7); (\dfrac{19}{3}, \dfrac{14}{3},\dfrac{19}{3})$
