• # question_answer Let a point R lies on the plane $x-y+z-3=0$ and P be the point (1, 1,1). A point Q lies on PR such that $P{{Q}^{2}}+\text{ }P{{R}^{2}}=k$ (constant), then the equation of locus of Q is A) $\left[ {{(x-1)}^{2}}+{{(y-1)}^{2}}+{{(z-1)}^{2}} \right]\left[ +\frac{4}{{{(x-y+z-1)}^{2}}} \right]=k$ B) $\left[ {{(x-1)}^{2}}+{{(y-1)}^{2}}+{{(z-1)}^{2}} \right]\left[ 1-\frac{4}{{{(x-y+z-1)}^{2}}} \right]=k$ C) ${{(x-1)}^{2}}+{{(y-1)}^{2}}+{{(z-1)}^{2}}+\frac{4}{{{(x-y+z-1)}^{2}}}=k$ D) $\frac{1}{{{(z-1)}^{2}}}+\frac{1}{{{(y-1)}^{2}}}+\frac{1}{{{(z-1)}^{2}}}+\frac{{{(x-y+z-1)}^{2}}}{4}=k$

Let Q be$(\alpha ,\beta ,\gamma )$then $P{{Q}^{2}}={{(\alpha -1)}^{2}}+{{(\beta -1)}^{2}}+{{(\gamma -1)}^{2}}=r_{2}^{2}$where$PQ={{r}_{2}}$ If $PR={{r}_{1}}$ and $l,m,n$ be the direction cosines of the line PR, then R is $(1+l{{r}_{1}},1+m{{r}_{1}},1+n{{r}_{1}})$ R lies on the plane, so${{r}_{1}}=\frac{2}{l-m+n}$ Also, Q is$(1+l{{r}_{2}},1=m{{r}_{2}}.1+n{{r}_{2}})$ $\Rightarrow$$\frac{\alpha -1}{{{r}_{2}}}=l,\frac{\beta -1}{{{r}_{2}}}=m,\frac{\gamma -1}{{{r}_{2}}}=n$ $\Rightarrow$${{r}_{1}}=\frac{2{{r}_{2}}}{\alpha -\beta +\gamma -1}$ Now,$r_{1}^{2}+r_{2}^{2}=k$ $\Rightarrow$$r_{2}^{2}\left[ 1+\frac{4}{(\alpha -\beta +\gamma -1)} \right]=k$ Locus of Q is $\left[ {{(x-1)}^{2}}+{{(y-1)}^{2}}+{{(z-1)}^{2}} \right]\left[ 1+\frac{4}{{{(x-y+z-1)}^{2}}} \right]=k$