Find the equation of plane passing through (3,2,0)and contain the line (x-3/1,y-6/5,z-4/4)

							
Let the equation of a plane passing through (3,2,0) bea(x−3)+b(y−2)+c(z−0)=0   ..(1)The line x−31=y−65=z−44passes through the point (3,2,0) and has direction ratios 1,5,4.If (1) contains the lines, it must pass through (3,6,4) and must be parallelto the line.Therefore,a(3−3)+b(6−2)+c(4−0)=0⇒4b+4c=0⇒b+c=0     ...(2)and a+5b+4c=0     ...(3)On solving (2) and (3), we geta=c and b=−cSo, From eq (1), we getc(x−3)−c(y−2)+c(z−0)=0⇒c[(x−3)−(y−2)+z]=0⇒(x−3)−(y−2)+z=0⇒x−y+z−1=0⇒x−y+z=1 is the required equation of plane.
						

							
Let the equation of plane passing through (3,2,0) is
a(x-3) + b(y-2) + c(z-0) = 0
 
x-3/1 = y-6/5 = z-4/4
Pass through (3,2,0) and has direction ratios as 1,5,4
 
If the equation of plane contains this, it should pass from
(3,6,4) and parallel to (1,5,4)
Hence
a(3-3) + b(6-2) + c *4 = 0
4b + 4c = 0
b +c = 0
And 
a + 5b +4 c = 0
 
On solving these two equations
a = c & b= -c
Hence equation of plane will be
c(x-3) - c(y-2) +cz = 0
x-3 -y +2 +z= 0
x - y +z -1 = 0
x -y +z = 1