The line frac{x-2}{3}=frac{y-3}{4}=frac{z-4}{ | vedclass

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(D) The direction ratios of the given line are $(3, 4, 5)$.For a line to be parallel to a plane,the normal vector of the plane must be perpendicular t

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$2x+y-2z=0$

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(D) The direction ratios of the given line are $(3, 4, 5)$.For a line to be parallel to a plane,the normal vector of the plane must be perpendicular to the direction vector of the line.If the plane equation is $ax+by+cz=d$,the normal vector is $(a, b, c)$.The condition for perpendicularity is $a_1a_2 + b_1b_2 + c_1c_2 = 0$.Checking option $D$: $2x+y-2z=0$,the normal vector is $(2, 1, -2)$.Calculating the dot product: $3(2) + 4(1) + 5(-2) = 6 + 4 - 10 = 0$.Since the dot product is $0$,the line is parallel to the plane $2x+y-2z=0$.

The line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is parallel to the plane

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