Planes, line and sphere in space

Problem

Given: the plane $\mu:\,-4x+2z+12=0$; the plane $\eta$ through $A(-6,-4,-6)$, $B(3,0,1)$, $C(-4,-1,-1)$; the sphere $\varepsilon:\,2x^{2}+2y^{2}+2z^{2}+16z-17=0$. a) Find the equation of $\eta$ and the angle between $\mu$ and $\eta$. b) Find the intersection line $\mu\cap\eta$. c) Find centre and radius of the sphere. d) Find the radius of the circle $\mu\cap\varepsilon$ and the planes parallel to $\mu$ tangent to the sphere.

Solution

a) Normal of $\eta$: $\vec{AB}\times\vec{AC}=(9,4,7)\times(2,3,5)=(-1,-31,19)$, so $\eta:\,-x-31y+19z-16=0$. With $\vec{n}_\mu=(-4,0,2)$: $\cos\alpha=\dfrac{|42|}{\sqrt{20}\,\sqrt{1323}}\approx0{,}258\Rightarrow\alpha\approx75{,}04^\circ$. b) From $\mu$: $z=2x-6$; setting $z=t$: $x=\tfrac{t}{2}+3$, $y=\tfrac{37t}{62}-\tfrac{19}{31}$, $z=t$. c) Dividing by $2$: $x^{2}+y^{2}+z^{2}+8z-\tfrac{17}{2}=0\Rightarrow x^{2}+y^{2}+(z+4)^{2}=\tfrac{17}{2}+16=\tfrac{49}{2}$. Centre $C_\varepsilon=(0,0,-4)$, radius $R=\sqrt{24{,}5}\approx4{,}95$. (The test’s solution stated $R=7$: it is wrong for this version — the constant $-17$ gives $R^{2}=24{,}5$, not $49$.) d) $d(C_\varepsilon,\mu)=\dfrac{|2(-4)+12|}{\sqrt{20}}=\dfrac{4}{\sqrt{20}}\approx0{,}894$, so $r=\sqrt{R^{2}-d^{2}}=\sqrt{24{,}5-0{,}8}\approx4{,}87$. Tangent planes $-4x+2z+k=0$ with $\dfrac{|k-8|}{\sqrt{20}}=R\Rightarrow k=8\pm\sqrt{490}$, i.e. $k\approx30{,}1$ or $k\approx-14{,}1$. $\boxed{\,\eta:-x-31y+19z-16=0,\ \ \alpha\approx75{,}04^\circ,\ \ C_\varepsilon(0,0,-4),\ R\approx4{,}95,\ r\approx4{,}87\,}$