algebraic representation of a line in 3d Is an algebraic representation of a line in 3d possible, or there can be only a parametric one? 
 A: In order to algebraically define a line you must algebraically define two planes and then take the intersection of these two planes.
We can define a line by taking
$$ (Ax + By + Cz = D) \wedge (A'x + B'y + C'z = D')$$
For valid constants. The planes must intersect but must not coincide. The symbol $\wedge$ means the values of $x,y,$ and $z$ must satisfy each equation.

For a generalization, consider working in $n$ dimensional space and try and find an algebraic representation of $k$-flats. A $k$ flat is simply a subset of the $n$ dimensional space that is congruent to Euclidian space of dimension $k$ for $k < n$. This means that $0$-flats are points, $1$-flats are lines, etc. The highest dimension flat is $n-1$ which is called a hyperplane. Your question involves finding and algebraic representation for $2$-flats in $3$ dimensional space.
Let $F_j$ denote an algebraic representation of some $j$ flat. We can now write any given $k$ flat for some set of $F_j$s as follows:
$$(F_{n-1}\wedge F_{n-1})\wedge F_{n-2} \wedge F_{n-3} \wedge \ldots \wedge F_{k+1}$$
Intuitively, we are intersecting two $(n-1)$-flats to get a $(n-2)$-flat and then intersection two $(n-2)$-flats to get a $(n-3)$-flat all the way until we are left intersecting two $(k+1)$-flats to get a $k$-flat. This is essentially a recursive definition because,  the way to define an $(n-2)$-flats would be to intersect some two $(n-1)$-flats. If you want to prove this representation, use induction.
An alternate definition would be to intersect some number of $(n-1)$-flats until you are left with a $k$-flat. This is equivalent to the first definition but not defined recursively. I prefer the first definition but that is up to the user.
In other words, if you want to define a point in $3$ dimensional space, intersect two planes and a line or, just intersect three planes. I prefer this method to other approaches because it algebraic (in my opinion) and generalization to any dimension of space and flat.

NB: This answer is essentially equivalent to jesterII's answer because his answer is simply another way to represent the intersection of two planes.
The equations
$$\frac{x-x_0}{a} =  \frac{y-y_0}{b}= \frac{z-z_0}{c}$$
Simply mean
$$\left(\frac{x-x_0}{a} =  \frac{y-y_0}{b} \right) \wedge \left(\frac{y-y_0}{b}=\frac{z-z_0}{c}\right)$$
It can be seen that in $3$ dimensional space both of the terms inside the parenthesis are planes which are then intersected to find the line.
A: Suppose we have described our line $\vec{r}$ in 3D using parametric form 
$$\vec{r}(t) = r_0 + t\vec{v}$$
where $t$ is our parameter.
It follows that $\vec{r}(t) = \langle x(t),y(t),z(t) \rangle$ is the position at time $t$, $r_0 = \langle x_0,y_0,z_0 \rangle$ is some point on the line, and  $\vec{v}=\langle a,b,c \rangle$ is a vector in the direction of the line.
Equating the left and right-hand side components of the equation, we find that
$$x(t) = x_0 + at$$
$$y(t) = y_0 + bt$$
$$z(t) = z_0 + ct$$
We can then eliminate the parameter $t$, which gives the following Cartesian representation of the line
$$ t = \frac{x-x_0}{a} =  \frac{y-y_0}{b}= \frac{z-z_0}{c}$$
A: I think the question being asked is, "Can the graph of $z=f(x,y)$ look like a line if $f$ has a nice simple expression?". I think one simple question to ask yourself is:


*

*What would the domain of such a function $f$ look like? 


Let's just do an example. Suppose $(0,0)$ and $(1,1)$ are in the domain of $f$, and suppose that above those points are $z$-values of $2$ and $3$, so that my graph contains the points $(0,0,2)$ and $(1,1,3)$. You know at least one way to write down lines in 3D, so you know some parametric equations of the line through these two points in 3D are $x=t$, $y=t$ and $z=2+t$. Notice that $x$ and $y$ are the same. So the entire domain of our function is the line $y=x$. In general, this sort of argument will show you that the domain of such a function is some line $y=mx+b$ in the $xy$-plane.
Okay, so our graph is a line, every point on our graph looks like $(x,y,f(x,y))$, and now we know that the only $x$ and $y$ values we can plug into $f$ lie on some line $y=x$. So really every point on our graph is $(x,x,f(x,x))$. So everything depends on $x$, and this is really two equations, $y = x$, and $z = f(x,x)$. In the general case, as in jesterII's comment, we will have $y=mx+b$ and $z=f(x,mx+b)$, where still everything depends on $x$. In this case, 
$x$ is nothing more than a parameter.
To convince yourself that you can't write a "simple" function $f(x,y)$ whose graph is a line, try writing down a function whose domain is the line $y=x$. 
A: The 3d space can be embedded projectively in an ambient 4d space.  Identify points in 3d space $r = x e_1 + y e_2 + z e_3$ with the projective points $p = r + e_4$ in the 4d ambient space.
Now, we can distinguish between points in the offset 3d flat and vectors which represent directions in that 3d flat.  In particular, a point $p$ will have a nonzero $e_4$ component, while a direction $v$ will have zero $e_4$ component.
Once this is done, exterior algebra allows us to represent any given line in the 3d flat using the wedge product.  Given two points $p,q$, the flat $p \wedge q$ corresponds to the line that they lie upon.  Equivalently, one could construct the difference vector $d = q-p$ and see that
$$p \wedge d = p \wedge (q - p) = p \wedge q - p \wedge p$$
But in exterior algebra, $p \wedge p = 0$, so this is entirely equivalent to the above expression:  you can wedge two points together, or you can wedge a point with the direction of the line going through it.
Lines represented this way are not unique, in the sense that any scalar multiple of the flat will represent the same line.  But this is customary in this kind of projective geometry: any scalar multiple of the 4d representation of a point is said to represent the same point in the 3d flat, too.
A: If you're asking "Can a (straight) line's equations be written without parameter t?", the general answer is yes. As already described in earlier posts, a straight line in 3-D can be regarded as the points P(x,y,z) which simultaneously satisfy equations for two planes, i.e. where those planes cross -- but note that two planes might not cross or might coincide, in which case there is no straight line "solution". What follows is a consideration of two plane equations that cross and produce a straight line (however, a straight line in 3-D space does not need to be defined as the crossing of two planes, in which case you may wish to jump to the third paragraph of this post).
To illustrate with an example, the equations for planes 3x + 2y + z = 1 and x + y - z = 0 do indeed meet along a line (see the Figure), but how to write it algebraically in a manner I think you're asking for? If possible, one obtains equations for y(x) and z(x) by solving the simultaneous equations, i.e. taking the two plane equations given and eliminating either z or y to get y(x) and z(x), respectively; in this example, the co-ordinate P(x,y,z) of any point along the straight line is given by:
$x$
$y = \frac{1}{3} - \frac{4}{3}x$
$z = \frac{1}{3} - \frac{1}{3}x$
One can now plug in any value of x to get its corresponding y- and z- co-ordinates for a point P(x,y,z) along the line; in the Figure, A and B are just two such points, i.e. when x = 0 and x = 1, respectively (red = x-axis, green = y-axis, blue = z-axis, graphics courtesy of the free on-line tool Geogebra here). At this point, one can move beyond the idea of a line needing to be the crossing of two planes and simply note that, in general, if x is one's independent variable, the equations for a point P(x,y,z) of a straight line in 3-D space will be of the following form (i.e. as described by @jll above):
$x$
$y = mx + c$
$z = nx + d$

