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 :
Dr?s of AB are (1, 2, 4) and dr?s of AC are (-2,-1,1) dr?s of normal to the plane ABC are (2,-3,1) Equation of plane ABC is \[2(x-1)-3(y-1)+1(z-1)=0\] or\[2x-3y+z=0\] Let required plane is \[2x-3y+z+k=0\] then\[\left| \frac{k}{\sqrt{4}+9+1} \right|=2\]or\[k=\pm 2\sqrt{14}\]You need to login to perform this action.
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