Question

Find the number of points of intersection of the plane $x+y+z=5\sqrt{3}$ and sphere $x^2+y^2+z^2=5$

Answer 1

Given sphere equation is $x^2+y^2+z^2=5$

Plane equation is $x+y+z=5\sqrt{3}$

Centre of sphere is $(0,0,0)$ and radius $\sqrt{5}$

distance from centre to plane

$=\frac{5\sqrt{3}}{\sqrt{1+1+1}}=\frac{5\sqrt{3}}{\sqrt{3}}=5$

as radius $\propto$ distance of centre to plane, there are no intersection points as line or plane lies outside sphere.