The equation of the plane through the points (2,-1,0), (3,-4,5) parallel to a line with direction cosines proportional to 2,3,4 is 9x-2y-3z=k, where k is

Options

A. 20
B. `-20`
C. 10
D. `-10`

Answer

A

Explanation

Equation of plane through (2,-1,0) is
`a(x-2)+b(y+1)+c(z-0)=0` . . . . (i)
It also passes through (3,-4,5), then
`a-3b+5c=0` . . . (ii)
Given plane (i) is parallel to a line with direction cosines proportional to 2,3,4.
`:." "2a+3b+4c=0" "[becausea_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0]` . . . (iii) From Eqs. (ii) and (iii), we get
`rArr" "(a)/(-12-15)=(b)/(10-4)=(c)/(3+6)rArr(1)/(-27)=(b)/(6)=(c)/(9)`
`rArr" "(a)/(9)=(b)/(-2)=(c)/(-3)=lamda" "` [say]
`:." "a=9lamda,b=-2lamda,c=-3lamda`
Now, from Eq. (i),
`9lamda(x-2)-2lamda(y+1)-3lamda(z=0)=0`
`rArr" "9x-2y-3z=20`
`:." "k=20`