2. ## Re: lines

You can equate x's, y's and z's and get three equations and three variables. Note that the value of t of the two lines don't have to be the same at the intersection point.

The vector (2, 1, 4) lies in the first line, and (a, -2, 1) lies in the second line. Once you know the intersection point and two vectors in the plane, you can use these methods to construct the plane equation.

3. ## Re: lines

Originally Posted by Kiiefers
$\left\{ {\begin{array}{l} {x = 1 + 2t} \\ {y = 7 + t} \\ {z = 3 + 4t} \\\end{array}} \right.\;\& \,\left\{ {\begin{array}{l} {x = 6 + as} \\ {y = - 1 - 2s} \\ {z = - 2 + s} \\\end{array}} \right.$
Then find the value of $a.$