Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The sphere $x^2+y^2+z^2=49$ has centre at the origin $(0,0,0)$ and radius 7.
Disance of the plane $2x+3y-z-5\sqrt{14}=0$ from the origin $$\begin{array}{l}=\frac{|2(0)+3(0)-(0)-5\sqrt{14}|}{\sqrt{2^2+3^2+(-1{)}^2}}\\ =\frac{|-5\sqrt{14}|}{\sqrt{14}}=\frac{5\sqrt{14}}{\sqrt{14}}=5\end{array}$$ Thus in Figure; $\mathrm{OP}=7,\mathrm{ON}=5$ $$\begin{array}{rl}{\mathrm{NP}}^2& ={\mathrm{OP}}^2-{\mathrm{ON}}^2=(7{)}^2-(5{)}^2\\ & =49-25=24\end{array}$$ $\therefore NP=2\sqrt{6}$ Hence the radius of the circle $$=\mathrm{NP}=2\sqrt{6}$$