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(C) The equation of the plane passing through $A(3, -2, 1)$ with normal vector $\vec{n} = 4\hat{i} + 7\hat{j} - 4\hat{k}$ is given by $\vec{n} \cdot (

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$\frac{28}{9}$

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(C) The equation of the plane passing through $A(3, -2, 1)$ with normal vector $\vec{n} = 4\hat{i} + 7\hat{j} - 4\hat{k}$ is given by $\vec{n} \cdot (\vec{r} - \vec{a}) = 0$. Substituting the values,we get $4(x - 3) + 7(y + 2) - 4(z - 1) = 0$. Simplifying this,$4x - 12 + 7y + 14 - 4z + 4 = 0$,which gives $4x + 7y - 4z + 6 = 0$. The length of the perpendicular from a point $P(x_1, y_1, z_1)$ to the plane $ax + by + cz + d = 0$ is given by $L = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$. Here,$P = (1, 2, -1)$ and the plane is $4x + 7y - 4z + 6 = 0$. Substituting these values,$L = \frac{|4(1) + 7(2) - 4(-1) + 6|}{\sqrt{4^2 + 7^2 + (-4)^2}}$. $L = \frac{|4 + 14 + 4 + 6|}{\sqrt{16 + 49 + 16}} = \frac{|28|}{\sqrt{81}} = \frac{28}{9}$. Thus,the length of $PM$ is $\frac{28}{9}$ units.

If $M$ is the foot of the perpendicular drawn from $P(1, 2, -1)$ to the plane passing through the point $A(3, -2, 1)$ and perpendicular to the vector $\vec{n} = 4\hat{i} + 7\hat{j} - 4\hat{k}$,then the length of $PM$,in proper units,is

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