Your lines must intersect to determine a plane; skew lines don't define a plane. But if they do intersect, then produce 3 non-collinear points $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$, $(x_3,y_3,z_3)$. In general, you will have a planar equation of the form $z = ax + by + c$ (except for the special case of a vertical plane). Your coefficients are the 3 unknowns, and you have three points. Using these 3 points, you can solve for your coefficients and get an equation for your plane.
Since you have two different lines, you can choose any two points on one of the lines, and a third point on the other line; you do not need to use their point of intersection. For convenience, try taking as many of the coordinates to be zero as possible.
In practical applications, your data will not be this nice. You will have a number of points scattered throughout space, not all lying on the same plane. Every three of these determines a plane, so it's typical to form a triangular mesh by repeatedly applying the above method for each local triplet. It's also more common to form two vectors and take a cross product to get a normal to the plane, and to write it in point-normal form; you may want to look into that method after you're comfortable with the one I mentioned above, especially if you're already familiar with vectors and the cross product.
Update: Now that I see the actual situation you have, your question is not basic at all. In fact, you have almost no hope of getting what I think you want out of your data. Your interpolated guess in the second figure is not likely close to the truth, but it also not the worst thing you could do with your data. I don't know what algorithm you used (or your software used) to get the interpolation, but there are other ways to guess the value of your function at the missing points. You should read the various methods outlined at http://en.wikipedia.org/wiki/Multivariate_interpolation. This is not my area of expertise, so maybe someone else will weigh in, but I think that you can probably find a method there that produces results that you prefer to the interpolation you already have, and also with some parameters that let you fine-tune the results. I think you'll probably end up with, at best, a very bad guess for the missing data, especially if you want to interpolate far from either of the curves. You will also need to choose a discrete subset of your curves to employ most of the methods, but I think you probably already have a discrete dataset that you used to produce them in the first place.
some options in R would be
t=seq(0,10,0.01)
y=sin(t)+rnorm(length(t))
plot(t,y,cex=0.1)
# puts a loess smoother through the points
lines(loess.smooth(t,y),col=2)
library(mgcv)
# puts a spline through the points - read Simon Woods work to learn more about this
g<-gam(y~s(t),data=data.frame(t=t,y=y))
lines(t,g$fitted.values,col=3)
and for your recently added data
t=1:400
y=rep(NA,length(t))
y[10]<-30
y[111]<-100
y[171]<-128
y[181]<-86
y[201]<-42
y[211]<-44
y[281]<-39
y[321]<-59
y[341]<-20
y[351]<-4
library(mgcv)
g<-gam(y~s(t),data=data.frame(t=t,y=y))
lm<-predict.gam(g,data.frame(t=t,y=y))
plot(t,y,cex=0.5,pch=16,ylim=c(min(lm),max(lm)))
lines(t,lm,col='red')
Best Answer
Let's assume that you have an equation for the curve, let's say for example, $y = \sqrt{x}$. The this equation is what defines the relationship between $x$-values and $y$-values. Given a known $x$-value, you just have to plug that value in for $x$ in the equation to find the $y$-value. In the example $y = \sqrt{x}$, When $x = 100$, the $y$-value is $y = \sqrt{100} = 10$.
If you are trying to plot an unfamiliar function, you may want to organize your work into a table of sample $x$-values and $y$-values. List a few $x$-values, and then use your equation to determine the appropriate $y$-values. Here's an example for the equation $y = \sqrt{x}$.
First, choose some $x$-values: $$ \begin{array}{l|l} x & y = \sqrt{x} \\ \hline -1 & \\ 0 & \\ 1 & \\ 2 & \\ 3 & \\ 4 & \\ \end{array} $$
Then, find the $y$-values, based on the given equation: $$ \begin{array}{l|l} x & y = \sqrt{x} \\ \hline -1 & \sqrt{-1} = \textrm{not a real number}\\ 0 & \sqrt{0} = 0\\ 1 & \sqrt{1} = 1\\ 2 & \sqrt{2} \approx 1.414\\ 3 & \sqrt{3} \approx 1.732\\ 4 & \sqrt{4} \approx 2\\ \end{array} $$ This tells you that some points on the graph include $(0,0)$, $(1,1)$, $(2,1.414)$, $(3,1.732)$, $(2,2)$. In fact, there are infinitely-many points on the curve, be we can often get a good idea of the shape of a simple graph based on a few points.
Hope this helps!