Oh yes, I do know division by zero is not defined. But look at this problem:
Find the Cartesian equation of the line that passes through $(3,-2,-5)$ and $(3,-2,6)$.
And the answer is conveniently stated to be: $$\frac{(x-3)}{0} = \frac{(y+2)}{0} = \frac{(z+5)}{11}$$
What gives us the liberty to divide by zero?
Insert a value of $x$; $2$ for example. $2$ divided by zero is infinity. But $\frac{(z+5)}{11}$ will give us a real value!
Please help me clear this silly confusion.
On
Let me start with an analogy. There are a number of ways of finding the equation of a line on the plane given two points.
For example, given the points $(a,b)$ and $(c,d)$, there’s the point/slope formula, where we first calculate the slope $m$ (which is $m=\frac{d-b}{c-a}$) and then the equation is given by $$ y-b = m(x-a)$$ (or $y-d = m(x-c)$).
Or you can use the 2-point formula, which says that the line has equation $$(c-a)(y-b) = (d-b)(x-a).\qquad\qquad(1)$$ This is sometimes “simplified” to $y-b = \frac{d-b}{c-a}(x-a)$ so that you can see the connection with the point/slope formula.
Or there is the “intercepts” formula, where you try to figure out the values of the points $(0,h)$ and $(l,0)$ the line goes through, and use those.
But the thing is that not all methods work for all lines: they all work for lines that have nonzero slope; and all but the last one work for lines that have zero slope. But the only one that will give you the right answer when the line is vertical is $(1)$: if $a=c$, you have a vertical line and the equation is of the form $x=a$, which you can get from equation (1) since you will get $0(y-b)=(d-b)(x-a)$, which simplifies to $(d-b)(x-a)=0$, which is equivalent (since $d\neq b$) to $x-a=0$. The line is described simply by saying what the $x$ coordinate is, $x=a$, without any specification on $y$.
The same thing is happening with your points. Because both points have the same $x$-coordinate and the same $y$-coordinate, your line is parallel to the $z$-axis, and so it is described by simply saying the constant value of $x$ and the constant value of $y$, and not mentioning $z$. The equation you are trying to use can only be used for lines that do not lie on planes parallel to the coordinate planes, which is not the case here. It’s like trying to use the point-slope formula for a vertical line. The line is specified by the equations $x=3$, $y=-2$.
(Or, to use another analogy, it’s like trying to use the quadratic formula to solve an equation of degree $1$, say $0x^2 + 2x-4=0$, and then complaining that the formula is telling you to divide by $0$... wrong tool for the problem.)
On
The equation you got as an answer is just an - IMO - unfortunate different way of giving the vector equation of the line in parameter form.
The line in parameter form is:
$$\begin{pmatrix} x\\ y\\z \end{pmatrix} = \begin{pmatrix} 3\\ -2\\-5 \end{pmatrix}+ t\begin{pmatrix} 3 -3\\ -2-(-2)\\6-(-5) \end{pmatrix} = \begin{pmatrix} 3\\ -2\\-5 \end{pmatrix}+t\begin{pmatrix} \color{blue}{0}\\ \color{green}{0}\\ \color{orange}{11} \end{pmatrix} $$
Rearranging this gives
$$\begin{pmatrix} x-3\\ y+2\\z+5 \end{pmatrix} = t \begin{pmatrix} \color{blue}{0}\\ \color{green}{0}\\ \color{orange}{11} \end{pmatrix} $$
If all entries of the direction vector were different from zero, you could solve this vector equation coordinatewise for $t$ which then gives your form of the equation. Now, some people are so attracted to this form that they extend it formally also to direction vectors with zeros in some coordinates. This then leads to
$$\frac{x-3}{\color{blue}{0}} = \frac{y+2}{\color{green}{0}} =\frac{z+5}{\color{orange}{11}}$$
The more precise formula for the line that passes through $(a,b,c)$ and $(d,e,f)$ is that it is given by the three equations \begin{align*} (x-a)(e-b)&=(y-b)(d-a)\\ (x-a)(f-c)&=(z-c)(d-a)\\ (y-b)(f-c)&=(z-c)(e-b). \end{align*} When $d-a$ and $e-b$ and $f-c$ are all nonzero, these can be combined into the single pair of equations $$ \frac{x-a}{d-a} = \frac{y-b}{e-b} = \frac{z-c}{f-c}. $$ However, that convenience comes at a cost of loss of generality, as you've noted. Nevertheless, the formula above does give the correct answer here: \begin{align*} (x-3)((-2)-(-2))&=(y-(-2))(3-3)\\ (x-3)(6-(-5))&=(z-(-5))(3-3)\\ (y-(-2))(6-(-5))&=(z-(-5))((-2)-(-2)), \end{align*} which simplifies to the correct equations for this line: \begin{align*} 0&=0 \\ x&=3 \\ y&=-2. \end{align*}