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

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l}Givenplaneis2x-2y+4z+5=0andthegivenpointsis\left(1,\frac{3}{2},2\right)\\ D\text{'}ratiosofthenormaltotheplaneare2,-2,4\\ So,the\text{?}equationoflinepassingthrough\left(1,\frac{3}{2},2\right)andwhosed\text{'}ratiosareequaltothe\\ D\text{'}ratiosofthenormaltotheplanei.e.,2,-2,4is\frac{x-1}{2}=\frac{y-\frac{3}{2}}{-2}=\frac{z-2}{4}=\lambda \\ \therefore Anypointintheplaneis2\lambda +1,-2\lambda +\frac{3}{2},4\lambda +2\\ Since,thepointliesintheplane,then\\ 2\left(2\lambda +1\right)-2\left(-2\lambda +\frac{3}{2}\right)+4\left(4\lambda +2\right)+5=0\\ \Rightarrow 4\lambda +2+4\lambda -3+16\lambda +8+5=0\\ \Rightarrow 24\lambda +12=0\therefore \lambda =-\frac{1}{2}\\ So,thecoordinatesofthepointintheplaneare\\ 2\left(-\frac{1}{2}\right)+1,-2\left(-\frac{1}{2}\right)+\frac{3}{2},4\left(-\frac{1}{2}\right)+2i.e.,0,\frac{5}{2},0\\ Hence,thefootoftheperpendicularis\left(0,\frac{5}{2},0\right)andthe\\ requiredlength=\sqrt[]{{\left(1-0\right)}^2+{\left(\frac{3}{2}-\frac{5}{2}\right)}^2+{\left(2-0\right)}^2}\\ =\sqrt[]{1+1+4}=\sqrt[]{6}units\end{array}$