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(C) The equation of the plane is $4x - 3y + z + 13 = 0$. The normal vector to the plane is $\vec{n} = 4\hat{i} - 3\hat{j} + \hat{k}$.The line passing

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$3\sqrt{\frac{7}{2}}$

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(C) The equation of the plane is $4x - 3y + z + 13 = 0$. The normal vector to the plane is $\vec{n} = 4\hat{i} - 3\hat{j} + \hat{k}$.The line passing through the origin $O(0, 0, 0)$ and perpendicular to the plane has the direction ratios $(4, -3, 1)$.The equation of this line is $\frac{x}{4} = \frac{y}{-3} = \frac{z}{1} = k$.Thus,any point on this line is of the form $(4k, -3k, k)$.Since $Q$ is the foot of the perpendicular,it lies on the plane. Substituting the coordinates of $Q$ into the plane equation:$4(4k) - 3(-3k) + (k) + 13 = 0$$16k + 9k + k + 13 = 0$$26k = -13 \Rightarrow k = -\frac{1}{2}$.Therefore,the coordinates of $Q$ are $(4(-\frac{1}{2}), -3(-\frac{1}{2}), -\frac{1}{2}) = (-2, \frac{3}{2}, -\frac{1}{2})$.Given $R = (-1, 1, -6)$,the distance $QR$ is:$QR = \sqrt{(-1 - (-2))^2 + (1 - \frac{3}{2})^2 + (-6 - (-\frac{1}{2}))^2}$$QR = \sqrt{(1)^2 + (-\frac{1}{2})^2 + (-\frac{11}{2})^2}$$QR = \sqrt{1 + \frac{1}{4} + \frac{121}{4}}$$QR = \sqrt{\frac{4 + 1 + 121}{4}} = \sqrt{\frac{126}{4}} = \sqrt{\frac{63}{2}} = \sqrt{\frac{9 	imes 7}{2}} = 3\sqrt{\frac{7}{2}}$.

Let $Q$ be the foot of the perpendicular from the origin to the plane $4x - 3y + z + 13 = 0$ and $R$ be a point $(-1, 1, -6)$ on the plane. Then the length $QR$ is

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