If the distance of the point P(1, -2, 1) from | vedclass

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(A) The distance of point $P(x_1, y_1, z_1) = (1, -2, 1)$ from the plane $Ax + By + Cz - D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt

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$\left(\frac{8}{3}, \frac{4}{3}, -\frac{7}{3}\right)$

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(A) The distance of point $P(x_1, y_1, z_1) = (1, -2, 1)$ from the plane $Ax + By + Cz - D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}}$.Given $d = 5$ and the plane $x + 2y - 2z - \alpha = 0$, we have $5 = \frac{|1(1) + 2(-2) - 2(1) - \alpha|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|1 - 4 - 2 - \alpha|}{\sqrt{9}} = \frac{|-5 - \alpha|}{3}$.Since $\alpha > 0$, $|-5 - \alpha| = |5 + \alpha| = 5 + \alpha$. Thus, $5 = \frac{5 + \alpha}{3} \Rightarrow 15 = 5 + \alpha \Rightarrow \alpha = 10$.The equation of the plane is $x + 2y - 2z = 10$.The line passing through $P(1, -2, 1)$ and perpendicular to the plane has direction ratios $(1, 2, -2)$. Its equation is $\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 1}{-2} = k$.Any point on this line is $(k + 1, 2k - 2, -2k + 1)$.Since this point lies on the plane, $(k + 1) + 2(2k - 2) - 2(-2k + 1) = 10$.$k + 1 + 4k - 4 + 4k - 2 = 10 \Rightarrow 9k - 5 = 10 \Rightarrow 9k = 15 \Rightarrow k = \frac{5}{3}$.Substituting $k = \frac{5}{3}$ into the point coordinates: $x = \frac{5}{3} + 1 = \frac{8}{3}$, $y = 2(\frac{5}{3}) - 2 = \frac{4}{3}$, $z = -2(\frac{5}{3}) + 1 = -\frac{7}{3}$.The foot of the perpendicular is $\left(\frac{8}{3}, \frac{4}{3}, -\frac{7}{3}\right)$.

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$, where $\alpha > 0$, is $5$, then the foot of the perpendicular from $P$ to the plane is

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