Q134 / 225 · JEE Main 2004 · Mathematics

The intersection of the spheres $x^2+y^2+z^2+7 x-2 y-z=13$ and $x^2+y^2+z^2-3 x+3 y+4 z=8$ is the same as the intersection of one of the sphere and the plane

A
$x-y-z=1$
B
$x-2 y-z=1$
C
$x-y-2 z=1$
Correct answer
D
$2 x-y-z=1$

Solution

Required plane is $S_1-S_2=0$ where $S_1=x^2+y^2+z^2+7 x-2 y-z-13=0$ and $$ \begin{aligned} & S_2=x^2+y^2+z^2-3 x+3 y+4 z-8=0 \\ & \Rightarrow 2 x-y-z=1 \end{aligned} $$

Hence, the correct answer is option: 'D'.

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