The eleven postulates are sufficient to prove 3.11.
Lemma 1 A line and a point not on it, two different lines in a plane, or two parallel lines define a plane.
Two points on a line and a point not on it define a plane by #7. If two lines are different there's a point on the second that's not on the first (by #6), so by the first part they define a plane. By definition two parallel lines are different lines in a plane so define it by the second part.
Lemma 2 If $a,b,t$ are different coplanar lines and $a$ is parallel to $b$ and $t$ is not parallel to $a$ then $t$ is a transversal of $a$ and $b$.
By definition $t$ intersects $a$ so call the point of intersection $A$ defining an angle $\angle at\ne 0$ (by #3). Let $S$ be a point on $b$ then $SA$ defines a line $s$ (by #6) which is a transversal of $a$ and $b$ (by definition). Then $s$ cuts off angles $\angle sb=\angle sa$ (by #10) and $\angle st\ne \angle sa$ (by #4 because they are coincident), so $t$ is not parallel to $b$ by $\angle st\ne \angle sb$ and #10, and is a transversal (by definition).
Proposition If $a,b,c$ are different lines with $a$ parallel to $b$ and $b$ parallel to $c$ then $a$ is parallel to $c$.
If the lines are coplanar then let $t$ be a line intersecting $b$, then applying Lemma 2 twice it is a common traversal of $a,b,c$. By #10 $\angle ta=\angle tb=\angle tc$ and by #11 $a$ is parallel to $c$.
If the lines are not coplanar, then let $C$ be a point on $c$. By Lemma 1 $a$ and $b$ are in a plane $\pi_1$, $b$ and $c$ are in a different plane $\pi_2$, and $a$ and $C$ are in a plane $\pi_3$. By #9 $\pi_2$ and $\pi_3$ intersect in a line $l$ that contains $C$.
$l$ cannot intersect $b$ in any point $B$, otherwise $a$ and $B$ are in both $\pi_1$ and $\pi_3$, so $\pi_1\equiv\pi_3$ by Lemma 1, $b\equiv \pi_1\cap\pi_2\equiv\pi_3\cap\pi_2\equiv l$ which would require $C$ to be on $b$, contradicting that $b$ and $c$ are parallel. So $l$ does not intersect $b$ but it intersects $c$ at $C$. Since $b,c,l$ are coplanar in $\pi_2$, by Lemma 2 they cannot all be different, so $l\equiv c$.
$l$ cannot intersect $a$ in any point $A$, otherwise $b$ and $A$ are in both $\pi_1$ and $\pi_2$, so $\pi_1\equiv\pi_2$ by Lemma 1, contradicting that $a,b,c$ are not coplanar. Since $l\equiv c$ and $a$ are both in $\pi_3$ and do not intersect it follows that $a$ is parallel to $c$.
Create vectors pointing along each line by computing $(x_2,y_2,z_2)-(x_1,y_1,z_1)$ for both pairs. Make them into unit vectors by dividing them by their lengths. Call these two unit vectors $u$ and $v$.
Then you can use the inner product identity: $\langle u,v\rangle=\cos(\theta)$, where $\theta$ is the angle between the two vectors.
You want to create two small thresholds around 1 and -1. When the dot product is close to 1, this means that the vectors are very nearly pointing in the same direction, and when the dot product is nearly -1, they are very close to pointing in opposite directions. In both cases, they are "nearly parallel".
Best Answer
If you are following Euclid's axioms and postulates, you cannot move a line. You may draw through any point $Z$ a line $XY$ parallel to $AB$ by making $\angle EZX=\angle ZEB$. But then you have to prove the corresponding angles are equal. Given two parallel lines and a transversal, Euclid [I, 29] first proves the alternate interior angles equal using his Postulate 5 (for the first time). Then using the equality of vertical angles [I, 15] he proves the corresponding angles equal.
Perhaps Euclid's postulate that you can draw a circle with any point as center and any distance (line) as radius constitutes an isolated case of moving a line in Euclidean geometry. And Descartes later proposed that curves produced by instruments more complex than compasses should be allowed in geometry, provided they were generated by a motion, or series of motions, strictly determined, e.g. the hyperbola and conchoid. But a curve like the quadratrix was ruled out as "mechanical" not geometrical, since it requires two coordinated but independent movements for its generation.
The present case calls not for a new curve but the movement of a given straight line to a new place while keeping the same angle to a tranversal. How exactly is this done? Or do we simply postulate its possibility?