Question

Find the locus of a point, the sum of squares of whose distance from the planes: $x-z=0,x-2y+z=0$ and $x+y+z=0$ is $36$

• A. $x^2+y^2+z^2=36$
• B. ${2x}^2+y^2+z^2=36$
• C. $x^2+{2y}^2+z^2=36$
• D. None of these

Answer 1

Answer: (A)

$x^2+y^2+z^2=36$

Given, planes are $x-z=0,x-2y+z=0$ and $x+y+z=0$

Let the point whose locus is required be $P(\alpha ,\beta ,\gamma )$.

According to the question,

$\frac{{|\alpha -\gamma |}^2}{2}+\frac{{|\alpha -2\beta +\gamma |}^2}{6}+\frac{{|\alpha +\beta +\gamma |}^2}{3}=36\Rightarrow 3({\alpha }^2+{\beta }^2+{\gamma }^2)+{\alpha }^2+4{\beta }^2+{\gamma }^2-4\alpha \beta -4\beta \gamma +2\alpha \gamma +2({\alpha }^2+{\beta }^2+{\gamma }^2+2\alpha \beta +2\beta \gamma +2\alpha \gamma )=36\times 6\Rightarrow 6{\alpha }^2+6{\beta }^2+6{\gamma }^2=36\times 6\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=36$

Hence, the equation of required locus is $x^2+y^2+z^2=36$