It will take the heck out of you!!!

The equation of sphere can be re-written as $(x-1)^2+(y+2)^2+z^2=5^2$

Therefore, centre of sphere is $(1,-2,0)$ and radius is $5$ .

The distance of point $(x_0,y_0,z_0)$ from a plane $ax+by+cz+d=0$ can be calculated by $\left |\dfrac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}\right |$

Substituting the values, we get distance between centre of sphere and plane is $\sqrt{11}$ .

Applying Pythagoras theorem, $5^2=r^2+\sqrt{11}^2$ . Solving, $r=\sqrt{14}≈3.74$ .

Centre of the sphere is $(1,-2,0)$ and its radius is 5 units. And distance of point $(1,-2,0)$ from plane $x + y + 3z -10 = 0$ is $\sqrt{11}$ units. So by Pythagoras Theorem, radius of circle is $\sqrt{5^2 - 11} = \sqrt{14} = 3.7416$ units.

Highly Over-rated Question

Find distance of centre from plane and apply Pythagoras to get radius of cross section.