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The foot of the perpendicular drawn from the origin to a plane is the point $(1,-3,1)$. What is the intercept cut on the $x$ -axis by the plane? $\quad$
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The correct answer is:
11
Equation of plane passing through $(1,-3,1)$ and whose normal $(1,-3,1)$ is $1(x-1)-3(y+3)+1(z-1)=0$
$\Rightarrow x-3 y+z-11=0$
$\Rightarrow \frac{x}{11}-\frac{y}{11 / 3}+\frac{z}{11}=0$
The above plane intercept the $x$ -axis at 11 .
$\Rightarrow x-3 y+z-11=0$
$\Rightarrow \frac{x}{11}-\frac{y}{11 / 3}+\frac{z}{11}=0$
The above plane intercept the $x$ -axis at 11 .
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