A plane passes through (2,3,-1) and is perpen | vedclass

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(D) The equation of a plane passing through the point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0

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$\frac{13}{\sqrt{74}}$

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(D) The equation of a plane passing through the point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.Given the point $(2, 3, -1)$,the equation is $a(x-2) + b(y-3) + c(z+1) = 0 \dots (i)$.Since the plane is perpendicular to the line with direction ratios $(3, -4, 7)$,the normal vector of the plane is parallel to this line.Thus,we can take $a=3, b=-4, c=7$.Substituting these into equation $(i)$:$3(x-2) - 4(y-3) + 7(z+1) = 0$$3x - 6 - 4y + 12 + 7z + 7 = 0$$3x - 4y + 7z + 13 = 0$.The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.Here,$A=3, B=-4, C=7, D=13$.$d = \frac{|13|}{\sqrt{3^2 + (-4)^2 + 7^2}} = \frac{13}{\sqrt{9 + 16 + 49}} = \frac{13}{\sqrt{74}}$.

$A$ plane passes through $(2,3,-1)$ and is perpendicular to the line having direction ratios $3,-4,7$. The perpendicular distance from the origin to this plane is

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