Problem: Given the line $$ passing through the two points $(1; -2; 0)$ and $(2; 3; -1)$, determine the equation Cartesian of the spherical surface of center $(1; -6; 7)$ and tangent to $$.
Partial solution: the line $r$ in parametric form is: $$\mathbf{v}=(l,m,n)=(x_B-x_A,y_B-y_A,z_B-z_A)$$
$$\begin{cases}x=x_A+lt=1+t \\ y=y_A+mt=-2+5t \\ z=z_A+nt=-t \end{cases}$$ with $t\in \Bbb R$.
Since the spherical surface is tangent to the line $$, its radius $$ will be equal to the distance between its center $(1; -6; 7)$ and the line $$. To calculate the distance between $$ and $$, we need to determine the equation of the plane $$ passing through $$ and perpendicular to the line $$.
Why the plane $$ use the directorial numbers $l,m,n$? I.e.
$$l(x-1)+m(y+6)+n(z-7)=0, \quad ?$$
Claim 1. Suppose you are given a vector $a=(a_1,a_2,a_3) \in \mathbb{R}^3$. Then, the plane of vectors perpendicular to $a$ is the set of vectors defined by $$ \{ b=(b_1,b_2,b_3) : a_1b_1+a_2b_2+a_3b_3=0 \}. $$ Proof. Let $|a|$ and $|b|$ be the lengths of the vector $a$ and $b$, respectively. Then, by the Pythagorean theorem, $a$ and $b$ are perpendicular if and only if \begin{align*} |a|^2+|b|^2 &= |a-b|^2 \quad\Longleftrightarrow \\ a_1^2+a_2^2+a_3^2 + b_1^2+b_2^2+b_3^2 &= (a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2 \quad\Longleftrightarrow \\ a_1^2+a_2^2+a_3^2 + b_1^2+b_2^2+b_3^2 &= a_1^2+b_1^2-2a_1b_1+a_2^2+b_2^2-2a_2b_2+a_3^2+b_3^2-2a_3b_3 \quad\Longleftrightarrow \\ 0&=-2a_1b_1-2a_2b_2-2a_3b_3. \end{align*} Dividing by $-2$ gives the desired result.
Back to your question.
Using Claim 1, the plane of vectors $(x',y',z')$ perpendicular to $(l,m,n)$ is given by $$lx'+my'+nz'=0.$$ It is quickly verified that this plane passes through $(x',y',z')=(0,0,0)$. So, in a second step, we translate the plane to pass through $C$ by setting $(x,y,z) = (x',y',z')+(1,-6,7)$, which gives the result $$l(x-1)+m(y+6)+n(z-7)=0.$$