You're doing fine. Everything you've said so far has been correct.
You're trying to build a system of equations to use to solve for a, b, and c. The dot product condition gives you another equation to use. Let's list what we know.
You gave:
<br />
\begin{align*}<br />
x &= 1+t\\<br />
y &= 1-t\\<br />
z &= 2t<br />
\end{align*}<br />
And the line we're looking for is given by:
<br />
\begin{align*}<br />
x &= as\\<br />
y &= 1 + bs\\<br />
z &= 2 + cs<br />
\end{align}<br />
Notice on the last set of equations I used "s" instead of "t" for the parameter. Two lines intersect when their equations are equal. Although we want the two lines to intersect at some point, it's quite possible that they could intersect at "different times." Therefore, we need to use a different parameter for each line when we set them equal (intersecting).
Finally, the dot product condition gives us:
<br />
\begin{align*}<br />
\langle 1,-1,2 \rangle \cdot \langle a,b,c \rangle &= 0\\<br />
a - b + 2c &= 0<br />
\end{align}<br />
Set the two lines equal to each other (x = x, y = y, z = z). This satisfies the "intersection" condition. The dot product result satisfies the "perpendicular" condition. With those two conditions, you should have four equations and five unknowns (a, b, c, t, s). More unknowns than available equations -- how can we solve this?
You must realize that there is no unique solution for a, b, and c. Why? Consider your original line. We said its direction vector was <1,-1,2>. Couldn't we use the direction vector <2,-2,4> to describe the same line? Or <6,-6,12>? They all point in the same direction; they just have different magnitudes.
My point is, you can pick an arbitrary value for a, b, or c (say, let c = 2) and then you have a system of four equations and four unknowns, which you can solve to give you the equation of your line, along with the s and t where they intersect.
