If {P_1} and {P_2} are the lengths of the per | vedclass

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(B) The length of the perpendicular from a point $(x_1, y_1, z_1)$ to the plane $Ax + By + Cz + D = 0$ is given by $P = \frac{|Ax_1 + By_1 + Cz_1 + D|

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$7{P^2} - 23P + 16 = 0$

Vedclass · Answer

(B) The length of the perpendicular from a point $(x_1, y_1, z_1)$ to the plane $Ax + By + Cz + D = 0$ is given by $P = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.For the plane $3x - 6y + 2z + 11 = 0$,the denominator is $\sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.For point $(2, 3, 4)$,${P_1} = \frac{|3(2) - 6(3) + 2(4) + 11|}{7} = \frac{|6 - 18 + 8 + 11|}{7} = \frac{|7|}{7} = 1$.For point $(1, 1, 4)$,${P_2} = \frac{|3(1) - 6(1) + 2(4) + 11|}{7} = \frac{|3 - 6 + 8 + 11|}{7} = \frac{|16|}{7} = \frac{16}{7}$.The quadratic equation with roots ${P_1}$ and ${P_2}$ is given by $(P - P_1)(P - P_2) = 0$,which is $P^2 - (P_1 + P_2)P + P_1P_2 = 0$.Sum of roots: $P_1 + P_2 = 1 + \frac{16}{7} = \frac{23}{7}$.Product of roots: $P_1P_2 = 1 	imes \frac{16}{7} = \frac{16}{7}$.Substituting these into the equation: $P^2 - \frac{23}{7}P + \frac{16}{7} = 0$,which simplifies to $7P^2 - 23P + 16 = 0$.

If ${P_1}$ and ${P_2}$ are the lengths of the perpendiculars from the points $(2, 3, 4)$ and $(1, 1, 4)$ respectively to the plane $3x - 6y + 2z + 11 = 0$,then ${P_1}$ and ${P_2}$ are the roots of the equation:

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