That is, the intersection of plane y=0 and plane z=ax, where a is a constant.
Add another x=x, and you can get the direction vector of the straight line (1, 0, a) and the given direction vector of the straight line (3,2, 1).
If the two vectors are perpendicular, you can get a=-3. therefore .....
2. Obviously, the plane formed by the intersection of the required straight line and the given straight line will form an intersection line with the plane given by the topic, and this intersection line is parallel to the required straight line.
Given the normal vector of plane q=(3, -4, 1)
A given straight line passes through a fixed point (-1, 3,0), and the direction vector m = (1, 1, 2).
The searched straight line passes through a fixed point (-1, 0,4).
So two points determine the vector n=(0, -3, 4), and the plane normal vector k=m×n=( 10, -4, -3) formed by two straight lines.
Then the direction vector of the intersection of two planes (also the direction vector of a straight line) l=q×k=(- 16,-19, -28).
There's ... ps off-point method. I can only guarantee that the steps are correct, and the result looks strange ~ ~