Too happening

let the centre of sphere =O(0,0,0).
circle = x^2 +y^2 + z^2.
ACO is right angle triangle with AO=1 and AC=2 .
thus OC^2 = 5.
CBO is a right triangle with CO^2=5 and BO =1.
CB=2.
z coordinate of of b=-3/13.
the radius of the circle which the sphere intercepts on the plane parallel to II on which b lies.
=(sqrt160) /13.
thus AB^2= ((sqrt160)/13)^2+(-1- -3/13)^2
=260/169=20/13

$$ Let point $\,O$ be the center of sphere. Length of $\,BC=AC=2\,$ since $\Delta \text{ACO}$ and $\Delta \text{BCO}$ are congruent. Set point $\,C\,$ as the origin and point $\,A\,$ on $\,y-$ axis. Let $\,(a,b,c)$ be the coordinate of point $\,B$ . Then, we have $\,A(0,2,0)\,$ and $\,O(0,2,1)$ . The equation of sphere can be express as $$ $x^2+(y-2)^2+(z-1)^2=1.$ $$ The coordinate of point $\,B\,$ can be express as $$ $a^2+b^2+c^2=2^2.$ $$ Since point $\,B\,$ also lies on the sphere, it can also be written as $$ $a^2+(b-2)^2+(c-1)^2=1.$ $$ $c\,$ can be determined by using $$ $\sin \phi = \frac{c}{BC}\;\;\;\;\;\Rightarrow\;\;\;\;\;c=BC \sin \phi = \frac{10}{13}.$ $$ Thus, $\begin{aligned} a^2+b^2+\left(\frac{10}{13}\right)^2&=4&&&(1)\\ a^2+(b-2)^2+\left(\frac{10}{13}-1\right)^2&=1.&&&(2) \end{aligned}$ $$ From (1) and (2), we obtain $\,a=\frac{3}{13}\sqrt{15}\,$ and $\,b=\frac{21}{13}$ . $AB^2$ can be obtain by using formula of distance between 2 points in Euclidean three-space. $$ $\begin{aligned} d^2&=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\\ AB^2&=\left(\frac{3}{13}\sqrt{15}-0\right)^2+\left(\frac{21}{13}-2\right)^2+\left(\frac{10}{13}-0\right)^2\\ &=\frac{20}{13}. \end{aligned}$ $$ Hence, $\,a+b=20+13=\boxed{33}$ . $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$