A) collinear
B) coplanar
C) non-coplanar
D) non-collinear
Correct Answer: B
Solution :
Equation of a plane through the point \[A(4,\,\,5,\,\,1)\]is \[a(x-4)+b(y-5)+c(z-1)=0\] ... (i) Points\[B(0,\,\,-1,\,\,-1)\]and\[C(3,\,\,9,\,\,4)\]lies on Eq. (i) \[\therefore \]\[a(0-4)+b(-1-5)+c(-1-1)=0\] \[\Rightarrow \] \[2a+3b+c=0\] ... (ii) and \[a(3-4)+b(9-5)+c(4-1)=0\] \[-a+4b+3c=0\] ... (iii) On solving Eqs. (ii) and (iii), we get \[\frac{a}{5}=\frac{b}{-7}=\frac{c}{11}\] On putting the values of \[a,\,\,\,b\] and \[c\] in Eq. (i), we get \[5(x-4)-7(y-5)+11(z-1)=0\] \[\Rightarrow \] \[5x-7y+11z+4=0\] Now, suppose \[D(-4,\,\,4,\,\,4)\] lies on it. \[\therefore \] \[5(-4)-7(4)+11(4)+4=0\] \[\Rightarrow \] \[0=0\] \[\therefore \]Points \[A,\,\,\,B,\,\,\,C\] and \[D\] are coplanar.
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