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A plane P passes through the line of intersection of the planes $2 x-y+3 z=2, x+y-z=1$ and the point $(1,0,1)$. If the plane $\mathrm{P}$ touches the sphere $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=\mathrm{r}^{2}$, then what is $r$ equal to?
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The correct answer is:
$\frac{5}{\sqrt{29}}$
Plane $\mathrm{P}$ touches the sphere $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=\mathrm{r}^{2}$ then
$\mathrm{r}=$ Distane between centre of sphere $(0,0,0)$ to plance $\mathrm{P}$.
$\Rightarrow \mathrm{r}=\left|\frac{5(0)+2(0)-5}{\sqrt{5^{2}+2^{2}+(0)^{2}}}\right|$
$\frac{5}{\sqrt{25+4}}$
$\mathrm{r}=$ Distane between centre of sphere $(0,0,0)$ to plance $\mathrm{P}$.
$\Rightarrow \mathrm{r}=\left|\frac{5(0)+2(0)-5}{\sqrt{5^{2}+2^{2}+(0)^{2}}}\right|$
$\frac{5}{\sqrt{25+4}}$
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