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A plane meets the coordinate axes at $P, Q, R$ respectively. If the centroid of $\triangle P Q R$ is $\left(1, \frac{1}{2}, \frac{1}{3}\right)$, then the equation of the plane is
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The correct answer is:
$x+2 y+3 z=3$
Let the equation of the plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Since, plane meets the coordinate axes at $P, Q, R$ respectively. So, the coordinates of the points $P, Q, R$ are $(a, 0,0),(0, b, 0),(0,0, c)$ respectively.
Also, centroid of $\triangle P Q R$ is $\left(1, \frac{1}{2}, \frac{1}{3}\right)$.
$$
\begin{array}{ll}
\therefore & \frac{a+0+0}{3}=1 \Rightarrow a=3 \\
& \frac{0+b+0}{3}=\frac{1}{2} \Rightarrow b=\frac{3}{2} \\
\text { and } & \frac{0+0+c}{3}=\frac{1}{3} \Rightarrow c=1
\end{array}
$$
$\therefore$ Equation of plane is $\frac{x}{3}+\frac{y}{3 / 2}+\frac{z}{1}=1$
$$
\Rightarrow \quad x+2 y+3 z=3
$$
Since, plane meets the coordinate axes at $P, Q, R$ respectively. So, the coordinates of the points $P, Q, R$ are $(a, 0,0),(0, b, 0),(0,0, c)$ respectively.
Also, centroid of $\triangle P Q R$ is $\left(1, \frac{1}{2}, \frac{1}{3}\right)$.
$$
\begin{array}{ll}
\therefore & \frac{a+0+0}{3}=1 \Rightarrow a=3 \\
& \frac{0+b+0}{3}=\frac{1}{2} \Rightarrow b=\frac{3}{2} \\
\text { and } & \frac{0+0+c}{3}=\frac{1}{3} \Rightarrow c=1
\end{array}
$$
$\therefore$ Equation of plane is $\frac{x}{3}+\frac{y}{3 / 2}+\frac{z}{1}=1$
$$
\Rightarrow \quad x+2 y+3 z=3
$$
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