$\text{ The length and foot of the }$$\text{ perpendicular from the point }$$\text{ (7, }14,5\text{ ) to the plane }$$2x+4y-z=2,\text{ are }$

$\text{ Let }M\text{ be the foot of perpendicular from }(7,14,5)\text{ to the given plane, then }$$PM\text{ is normal to the plane. So, its d.r.'s are }2,4,-1\text{ . Since }PM\text{ passes }$

$\text{ through }P(7,14,5)\text{ and has d.r.'s }2,4,-1$

$\begin{array}{l}\text{ Therefore, its equation is }\frac{x-7}{2}=\frac{y-14}{4}=\frac{z-5}{-1}=r\text{ , }\\ \Rightarrow x=2r+7,y=4r+14,z=-r+5\\ \text{ Co-ordinates of }M\text{ be }(2r+7,4r+14,-r+5)\\ \text{ Since M lies on the plane }2x+4y-z=2,\text{ therefore }\\ 2(2r+7)+4(4r+14)-(-r+5)=2\Rightarrow r=-3\\ \text{ Co-ordinates of foot of perpendicular are }M(1,2,8)\text{ . }\\ PM=\text{ Length of perpendicular from }P\\ =\sqrt{{(1-7)}^2+{(2-14)}^2+{(8-5)}^2}=3\sqrt{21}.\end{array}$