Question

Set two planes whose intersection is the parametric line x=2+t; y=3t; z=1-t (t belongs to R)

Answer 1

The line has a directional vector of <1,3,-1> that goes through point (2,0,1)
The 2 planes will be in this form:
ax +by +z = c
dx +ey +z = f
the normal vectors are <a,b,1> and <d,e,1> respectively
by taking the cross product of the normal vectors you can obtain the vector <1,3,-1>
also plugging in the point (2,0,1) into both plane equations for x,y,z
this yields the following 5 equations:
b-e = 1
a-d = -3
ae -bd = -1
2a +1 = c
2d +1 = f
since there are 6 variables, 1 is a free variable. Let f = 0
--> d = -1/2
--> a = -7/2
--> c = -6
this leaves
b-e = 1
-7/2e +1/2b = -1

--> e = 1/2
--> b = 3/2
So 2 possible planes (written in standard form) that intersect with given line are
-7x +3y +2z = -12
-x + y +2z = 0