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

Equation of the given plane is $2x - 2y + 4z + 5 = 0$

……(i)
$\Rightarrow$ $\overrightarrow {\rm{n}} = 2\widehat {\rm{i}} - 2\widehat {\rm{j}} + 4\widehat {\rm{k}}$

So, the equation of line through $\left( {1,\frac{3}{2},2} \right)$ and parallel to $\overrightarrow {\rm{n}}$ is given by

$\frac{{x - 1}}{2} = \frac{{y - 3/2}}{{ - 2}} = \frac{{z - 2}}{4} = \lambda$

$\Rightarrow$ $x = 2\lambda + 1,y = - 2\lambda + \frac{3}{2}$ and $z = 4\lambda + 2$

If this point lies on the given plane, then

$2(2\lambda + 1) - 2\left( { - 2\lambda + \frac{3}{2}} \right) + 4(4\lambda + 2) + 5 = 0$

[using Eq. (i)]
$\Rightarrow$ $4\lambda + 2 + 4\lambda - 3 + 16\lambda + 8 + 5 = 0$

$\Rightarrow$ $24\lambda = - 12 \Rightarrow \lambda = \frac{{ - 1}}{2}$
$\therefore$

Required foot of perpendicular $= \left[ {2 \times \left( {\frac{{ - 1}}{2}} \right) + 1, - 2 \times \left( {\frac{{ - 1}}{2}} \right) + \frac{3}{2},4 \times \left( {\frac{{ - 1}}{2}} \right) + 2} \right]$

i.e., $\left( {0,\frac{5}{2},0} \right)$
$\therefore$

Required length of perpendicular $= \sqrt {{{(1 - 0)}^2} + {{\left( {\frac{3}{2} - \frac{5}{2}} \right)}^2} + {{(2 - 0)}^2}}$

$= \sqrt {1 + 1 + 4} = \sqrt 6$ units