Show that the plane whose vector equation is 
 contains the line whose vector equation is ![]()
we know that plane 
 contains the line 
 if
![]()
![]()
Given, equation of plane 
 and equation of line ![]()
So ![]()
d = 3 ![]()
![]()
= (2)(1) + (1)(2) + (4)(– 1)
= 2 + 2 – 4
![]()
![]()
= (1)(1) + (1)(2) + (0)(– 1)
= 1 + 2 – 0
= 3
= d
Since, 
 and ![]()
So, Given line is on the given plane.
Hence, Proved.