Question

The pole of the plane $3x+2y-4_Z-11=0$ w.r.t sphere $x^2+y^2+z^2=33$ is

• A. $\left(\frac{9}{2},\frac{1}{4},3\right)$
• B. $(2,1,-6)$
• C. $(9,6,-12)$
• D. $\left(\frac{1}{2}\frac{-3}{4},\frac{5}{6}\right)$

Answer 1

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.
$\Rightarrow \frac{a}{3}=\frac{b}{2}=\frac{c}{-4}=\frac{33}{11}$
Hence, required pole is $(9,6,-12)$