Find the length and the foot of the perpendicular from the point $\left(1, \frac{3}{2}, 2\right)$ to the plane $2x - 2y + 4z + 5 = 0$.

(A) The equation of the given plane is $2x - 2y + 4z + 5 = 0$ ... $(i)$
The normal vector to the plane is $\vec{n} = 2\hat{i} - 2\hat{j} + 4\hat{k}$.
The equation of the line passing through the point $P\left(1, \frac{3}{2}, 2\right)$ and perpendicular to the plane is given by $\frac{x-1}{2} = \frac{y-3/2}{-2} = \frac{z-2}{4} = \lambda$.
Thus,any point on this line is given by $x = 2\lambda + 1$,$y = -2\lambda + \frac{3}{2}$,and $z = 4\lambda + 2$.
If this point lies on the plane,it must satisfy the plane equation:
$2(2\lambda + 1) - 2(-2\lambda + \frac{3}{2}) + 4(4\lambda + 2) + 5 = 0$
$4\lambda + 2 + 4\lambda - 3 + 16\lambda + 8 + 5 = 0$
$24\lambda + 12 = 0 \Rightarrow 24\lambda = -12 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ into the coordinates,the foot of the perpendicular is $\left(2(-\frac{1}{2}) + 1, -2(-\frac{1}{2}) + \frac{3}{2}, 4(-\frac{1}{2}) + 2\right) = \left(0, \frac{5}{2}, 0\right)$.
The length of the perpendicular is the distance between $\left(1, \frac{3}{2}, 2\right)$ and $\left(0, \frac{5}{2}, 0\right)$:
$d = \sqrt{(1-0)^2 + (\frac{3}{2} - \frac{5}{2})^2 + (2-0)^2} = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$ units.