The coordinates of the foot of the perpendicu | vedclass

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(D) The equation of the line passing through the origin $(0, 0, 0)$ and perpendicular to the plane $3x + 2y + 6z - 56 = 0$ is given by $\frac{x-0}{3}

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$\left(\frac{24}{7}, \frac{16}{7}, \frac{48}{7}\right)$

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(D) The equation of the line passing through the origin $(0, 0, 0)$ and perpendicular to the plane $3x + 2y + 6z - 56 = 0$ is given by $\frac{x-0}{3} = \frac{y-0}{2} = \frac{z-0}{6} = k$.Any point on this line is $(3k, 2k, 6k)$.Since this point lies on the plane,we substitute it into the plane equation:$3(3k) + 2(2k) + 6(6k) = 56$$9k + 4k + 36k = 56$$49k = 56$$k = \frac{56}{49} = \frac{8}{7}$Substituting $k$ back into the coordinates $(3k, 2k, 6k)$:$x = 3 	imes \frac{8}{7} = \frac{24}{7}$$y = 2 	imes \frac{8}{7} = \frac{16}{7}$$z = 6 	imes \frac{8}{7} = \frac{48}{7}$Thus,the foot of the perpendicular is $\left(\frac{24}{7}, \frac{16}{7}, \frac{48}{7}ight)$.

The coordinates of the foot of the perpendicular drawn from the origin to the plane $3x + 2y + 6z = 56$ are:

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