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

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$\frac{28}{9}$

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

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 $4 \hat{i}+7 \hat{j}-4 \hat{k}$, then the length of $PM$ is

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