A) \[2x-3y+z+3\sqrt{14}=0\]
B) \[2x-3y+z+2\sqrt{14}=0\]
C) \[2x-3y+z+\sqrt{14}=0\]
D) \[2x-3y+z-2\sqrt{14}=0\]
Correct Answer: A
Solution :
Equation of plane passing through the points \[(1,1,1)\] is \[a(x-1)+b(y-1)+c(z-1)=0\] ?.(i) If this plane passes through the points \[(2,3,5)\] and \[(-1,\,0,\,2)\] Then, \[a(-1-1)+b(0-1)+c(2-1)=0\] \[\Rightarrow \] \[-2a-b+c=0\] ?.(ii) and \[a(2-1)+b(3-1)+c(5-1)=0\] \[\Rightarrow \]\[a+2b+4c=0\] ?..(iii) On solving Eqs. (ii) and (iii), we get \[\frac{a}{-1\times 4-1\times 2}=\frac{b}{1\times 1-(-2)(4)}=\frac{c}{(-2)(2)-(-1)(1)}\]\[\Rightarrow \] \[\frac{a}{-6}=\frac{b}{9}=\frac{c}{-3}=k\] \[\Rightarrow \] \[a=-6k,\,\,b=9k,\,\,c=-3k\] On putting the values of a, b and c in Eqs. (i), we get \[(-6k)\,(x-1)+9k(y-1)+(-3k)(z-1)=0\] \[\Rightarrow \] \[(-6)(x-1)+9(y-1)+(-3)(z-1)=0\] \[\Rightarrow \] \[-6x+6+9y-9-3z+3=0\] \[\Rightarrow \] \[-6x+9y-3z=0\] \[\Rightarrow \] \[-2x+3y-z=0\] \[\Rightarrow \] \[2x-3y+z=0\] Any plane parallel to the plane \[2x-3y+z=0\]is \[2x-3y+z+\lambda =0\] Given distance between planes is 3. Let \[P({{x}_{1}},{{y}_{1}},{{z}_{1}})\] be any point on \[2x-3y+z=0\] Then, \[2{{x}_{1}}-3{{y}_{1}}+{{z}_{1}}=0\] Length of the perpendicular from \[P({{x}_{1}},{{y}_{1}},{{z}_{1}})\] to \[2x-3y+z+\lambda =0\] is \[\frac{|2{{x}_{1}}-3{{y}_{1}}+{{z}_{1}}+\lambda |}{\sqrt{{{(2)}^{2}}+{{(-3)}^{2}}+{{(1)}^{2}}}}=3\] \[\Rightarrow \] \[\frac{(2{{x}_{1}}-3{{y}_{1}}+{{z}_{1}})+\lambda }{\sqrt{4+9+1}}=3\] \[\Rightarrow \] \[\frac{0+\lambda }{\sqrt{14}}=\frac{\lambda }{\sqrt{14}}=3\] \[\Rightarrow \] \[\lambda =3\sqrt{14}\] \[\therefore \] Required equation of plane is \[2x-3y+z+3\sqrt{14}=0\]
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