A) \[x+y+z=1\]
B) \[x+y+z=2\]
C) \[x+y+z=0\]
D) None of the above
Correct Answer: D
Solution :
The equation of the 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 Eqs. (ii) and (iii), we get \[\frac{a}{-14}=\frac{b}{-14}=\frac{c}{-14}\]or\[\frac{a}{1}=\frac{b}{1}=\frac{c}{1}\] So, the required plane is\[~x+y+z=0\].You need to login to perform this action.
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