A) \[2x-3y+z+2\sqrt{14}=0\]
B) \[2x-3y+z-\sqrt{14}=0\]
C) \[2x-3y+z+2=0\]
D) \[2x-3y+z-2=0\]
Correct Answer: A
Solution :
[a] A (1, 1, 1), B(2, 3, 5), C (-1, 0, 2) direction ratios of AB are <1, 2, 4>. Direction ratios of AC are \[<-\,2,\,\,-1,\text{ }1>.\] Therefore, directions ratios of normal to plane ABC are \[<2,\,-3,\text{ }1>\] As a result, equation of the plane ABC is\[2x-3y+z=0\]. Let the equation of the required plane be \[2x-3y+z=k.\] Then \[\left| \frac{k}{\sqrt{4+9+1}} \right|=2\] Or \[k=\pm 2\sqrt{14}\] Hence, equation of the required plane is \[2x-3y+z+2\sqrt{14}=0\]
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