A) \[x+y+z=1\]
B) \[x+y+z=2\]
C) \[x+y+z=0\]
D) None of these
Correct Answer: C
Solution :
The equation of plane containing the line \[\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}\] is \[a(x+1)+b(y-3)+c(z+2)=0\] .....(i) where \[-3a+2b+c=0\] .....(ii) This passes through (0, 7, ?7) \[\therefore \] \[a+4b-5c=0\] ..?(iii) From (ii) and (iii),\[\frac{a}{-14}=\frac{b}{-14}=\frac{c}{-14}\] or \[\frac{a}{1}=\frac{b}{1}=\frac{c}{1}\] Thus, the required plane is \[x+y+z=0\].You need to login to perform this action.
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