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

Explanation

The equation of the plane containing line `(x+1)/(-3)=(y-3)/(2)=(z+2)/(1)` is
`" "a(x+1)+b(y-3)+c(z+2)=0" "`(i)
where `" "-3a+2b+c=0" "` (ii)
This passes through (0, 7, -7). Thus,
`" "a+4b-5c=0" "`(iii)
From (i) and (ii), `(a)/(-14)=( b)/(-14)=(c)/(-14)or (a)/(1)=(b)/(1)=(c)/(1)`
So, the required plane is `x+y+z=0`.