Answer:
[tex]x+y=-2[/tex] and [tex]3z-y=5[/tex] are two such planes.
Step-by-step explanation:
To find the two planes whose intersection is the line
[tex]x=6+3t\\ y=-8-3t\\ z=-1-t[/tex]
You can say that t is equal to this expression
[tex]t=\frac{x-6}{3} =\frac{y+8}{-3}=-z-1[/tex]
Next,
[tex]\frac{x-6}{3} =\frac{y+8}{-3}\\ \frac{y+8}{-3}=-z-1[/tex]
Then,
[tex]x+y=-2[/tex] and [tex]3z-y=5[/tex] are two such planes.
You can see in the image attached that the intersection of this planes is the line
[tex]x=6+3t\\ y=-8-3t\\ z=-1-t[/tex]
