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If the distance of the point $P(1,-2,1)$ from the plane $x+2 y-2 z=\alpha$, where $\alpha>0$, is 5 , then the foot of the perpendicular from $P$ to the plane is
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Verified Answer
The correct answer is:
$\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$
$\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$
$$
\text { Distance of point } P \text { from plane }=5
$$
$$
\begin{gathered}
\therefore=\left|\frac{1-4-2-\alpha}{3}\right| \\
\alpha=10
\end{gathered}
$$
Foot of perpendicular
$$
\begin{aligned}
\quad \frac{x-1}{1} & =\frac{y+2}{2}=\frac{z-1}{-2}=\frac{5}{3} \\
\Rightarrow \quad \quad \quad x & =\frac{8}{3}, y=\frac{4}{3}, z=-\frac{7}{3}
\end{aligned}
$$
Thus, the foot of the perpendicular is
$$
A\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)
$$
\text { Distance of point } P \text { from plane }=5
$$
$$
\begin{gathered}
\therefore=\left|\frac{1-4-2-\alpha}{3}\right| \\
\alpha=10
\end{gathered}
$$
Foot of perpendicular
$$
\begin{aligned}
\quad \frac{x-1}{1} & =\frac{y+2}{2}=\frac{z-1}{-2}=\frac{5}{3} \\
\Rightarrow \quad \quad \quad x & =\frac{8}{3}, y=\frac{4}{3}, z=-\frac{7}{3}
\end{aligned}
$$
Thus, the foot of the perpendicular is
$$
A\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)
$$
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