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? 

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.
