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If $(2,-1,3)$ is the foot of the perpendicular drawn from the origin to a plane, then the equation of that plane is
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Verified Answer
The correct answer is:
$2 x-y+3 z-14=0$
Let equation of plane be
$$
a x+b y+c z+d=0
$$
Where $a, b, c$ are direction ratios of line perpendicular to plane
$$
\begin{aligned}
& a=2-0=2, \\
& b=-1-0=-1 \\
& c=3-0=3 \\
& \therefore 2 x-y+3 z+\mathrm{d}=0
\end{aligned}
$$
$\because$ It passes through $(2,-1,3)$
$$
\begin{aligned}
& 2(2)-(-1)+3(3)+d=0 \\
& d=-14 \\
& \therefore 2 x-y+3 z-14=0
\end{aligned}
$$
$$
a x+b y+c z+d=0
$$
Where $a, b, c$ are direction ratios of line perpendicular to plane
$$
\begin{aligned}
& a=2-0=2, \\
& b=-1-0=-1 \\
& c=3-0=3 \\
& \therefore 2 x-y+3 z+\mathrm{d}=0
\end{aligned}
$$
$\because$ It passes through $(2,-1,3)$
$$
\begin{aligned}
& 2(2)-(-1)+3(3)+d=0 \\
& d=-14 \\
& \therefore 2 x-y+3 z-14=0
\end{aligned}
$$
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