The pole of the plane 3x+2y−4Z−11=0 w.r.t sphere x2+y2+z2=33 is
A
(92,14,3)
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B
(2,1,−6)
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C
(9,6,−12)
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D
(12−34,56)
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Solution
The correct option is C(9,6,−12) Let pole be P(a,b,c) So the plane corresponding to P is ax+by+cz−33= (i) Now the plane 3x+2y−4z−11=0 and (i) are same. ⇒a3=b2=c−4=3311 Hence, required pole is (9,6,−12)