Find ten equation of the plane passing through the point `(0,7,-7)` and containing the line `(x+1)/(-3)=(y-3)/2=(z+2)/1` .

Explanation

The general equation of the plane passing through the point (0,7,-7) is given by
`a(x-0)+b(y-7)=0
If (i) contains the given line then it must pass through the point *-1,3,-2) and must be parallel to the given line
If (i) passes through the point (-1,3,-2) we have
a(-1-0)+b(3-7)+c(-2+7)=0 rarr a+4b-5c=0
If (i) is parallel to the given line then its normal should be perpendicular to htis line
`therefore (-3) a+2b+1xxc=0 rarr -3a+2b+c=0`
on solving (ii) and (iii) by cross multiplication we get
`rarr (A)/(1)=(b)/(1)=(c )/(1) = lambda`(say)
Then `a=lambda, b = lambda` and `c = lambda`
Putting `a=lambda , b = lambda` and `c = lambda` in we get
`lambda x+lambda(y-7) + lambda(z+7) =0 rarr x + (y-7) + (z+7)=0`
`rarr x+y+z=0`
Hence the required equation of the plane is x + y Z=0