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(B) The given equation of the sphere is $x^{2} + y^{2} + z^{2} - 2y - 4z = 11$. Rewriting it in standard form: $x^{2} + (y-1)^{2} - 1 + (z-2)^{2} - 4

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$\sqrt{7}$

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(B) The given equation of the sphere is $x^{2} + y^{2} + z^{2} - 2y - 4z = 11$. Rewriting it in standard form: $x^{2} + (y-1)^{2} - 1 + (z-2)^{2} - 4 = 11$,which simplifies to $x^{2} + (y-1)^{2} + (z-2)^{2} = 16$. The center of the sphere is $C(0, 1, 2)$ and its radius $R = \sqrt{16} = 4$. The distance $d$ from the center $C(0, 1, 2)$ to the plane $x + 2y + 2z - 15 = 0$ is given by $d = \frac{|0 + 2(1) + 2(2) - 15|}{\sqrt{1^{2} + 2^{2} + 2^{2}}} = \frac{|2 + 4 - 15|}{\sqrt{1 + 4 + 4}} = \frac{|-9|}{3} = 3$. The radius $r$ of the circle formed by the intersection is given by $r = \sqrt{R^{2} - d^{2}}$. Substituting the values,$r = \sqrt{4^{2} - 3^{2}} = \sqrt{16 - 9} = \sqrt{7}$.

Find the radius of the circle formed by the intersection of the sphere $x^{2} + y^{2} + z^{2} - 2y - 4z = 11$ and the plane $x + 2y + 2z = 15$.

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