Find the planes tangent to x^2+y^2+z^2-10x+2y+26z-113=0 and…

Find the planes tangent to $x^2+y^2+z^2-10x+2y+26z-113=0$ and parallel to the lines $\frac{x+5}2=\frac{y-1}{-3}=\frac{z+13}2,\qquad \frac{x+7}3=\frac{y+1}{-2}=\frac{z-8}0 $.

Explanation of Method

Tangent planes to a sphere that are parallel to two given lines.

Step 1: Sphere

\[ (x-5)^2+(y+1)^2+(z+13)^2=113+25+1+169=308,\qquad C(5,-1,-13),\ R=2\sqrt{77}. \]

Step 2: Normal

\[ \vec n=\{2,-3,2\}\times\{3,-2,0\}=\{4,6,5\},\qquad |\vec n|=\sqrt{77}. \]

Step 3: Tangency

\[ \frac{|20-6-65+d|}{\sqrt{77}}=2\sqrt{77}\ \Longrightarrow\ |d-51|=154\ \Longrightarrow\ d=205\ \text{ or }\ d=-103. \]

\[4x+6y+5z+205=0\qquad\text{and}\qquad 4x+6y+5z-103=0\]

Neither plane contains either line: the points $(-5,1,-13)$ and $(-7,-1,8)$ do not satisfy the equations. So the lines are strictly parallel to both planes.

$\vec n$ $C$ $R=2\sqrt{77}$

$4x+6y+5z-103=0$
$4x+6y+5z+205=0$

Section along the normal $\vec n=\{4,6,5\}$: the two tangent planes are perpendicular to $\vec n$, at distance $R=2\sqrt{77}$ on either side of $C(5,-1,-13)$.

Result

\[\boxed{\,4x+6y+5z+205=0\qquad\text{and}\qquad 4x+6y+5z-103=0\,}\]