If for $\mathrm{a}>0$,the feet of perpendiculars from the point $\mathrm{A}(\mathrm{a},-2\mathrm{a},3)$ and $\mathrm{B}(0,4,5)$ on the plane $\mathrm{lx}+\mathrm{my}+\mathrm{nz}=0$ are points $\mathrm{C}(0,-\mathrm{a},-1)$ and $\mathrm{D}$ respectively, then the length of line segment $\mathrm{CD}$ is equal to:

The correct option is D

$\sqrt{66}$

Explantion for the correct option:

Step 1: Representation of given details in form of a diagram

We have to find the length of the CD.

Given, that the equation of the plane is $\mathrm{lx}+\mathrm{my}+\mathrm{nz}=0$

As $C$ lies on this plane $\mathrm{C}(0,-\mathrm{a},-1)$ should satisfy the plane equation.

Substituting, $\mathrm{x}=0,\mathrm{y}=-\mathrm{a},\mathrm{z}=-1$

$$\begin{gathered} \Rightarrow 0-\mathrm{am}-\mathrm{n}=0 \\ \Rightarrow -\mathrm{am}-\mathrm{n}=0 \\ \Rightarrow -\mathrm{am}=\mathrm{n} \\ \Rightarrow \mathrm{am}=-\mathrm{n} \\ \Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{-1}{\mathrm{a}}...\left(1\right) \end{gathered}$$

Step 2: Finding the direction ratio of $\mathrm{CA}$

The formula for Direction Ratio: $\left({\mathrm{x}}_2-{\mathrm{x}}_1,{\mathrm{y}}_2-{\mathrm{y}}_1,{\mathrm{z}}_2-{\mathrm{z}}_1\right)$

Direction ratio of $\mathrm{CA}=(\mathrm{a},-\mathrm{a},4)$

As the direction ratio of $CA$ is parallel to the direction ratio of plane

We can write it as,

$\frac{\mathrm{a}}{\mathrm{l}}=\frac{-\mathrm{a}}{\mathrm{m}}=\frac{4}{\mathrm{n}}$

$$\begin{gathered} \Rightarrow \frac{-\mathrm{a}}{\mathrm{m}}=\frac{4}{\mathrm{n}} \\ \Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{-\mathrm{a}}{4}...\left(2\right) \end{gathered}$$

Step 3: Finding the value of ‘a’

From equations (1) and (2),

$\begin{array}{rcl}\therefore \frac{-\mathrm{a}}{4}& =& \frac{-1}{\mathrm{a}}\\ & \Rightarrow & {\mathrm{a}}^2=4\\ & \Rightarrow & \mathrm{a}=\pm 2\\ & \Rightarrow & \mathrm{a}=2(\mathbf{\because }\mathbf{given}\mathbf{,}\mathbf{}\mathbf{a}\mathbf{>}\mathbf{0})\end{array}$

Therefore, substituting the value of ‘$a$’ in equation (1), $\frac{\mathrm{m}}{\mathrm{n}}=\frac{-1}{2}$

Let us assume

$\begin{array}{rcl}& \Rightarrow & \mathrm{m}=-\mathrm{t}\\ & \Rightarrow & \mathrm{n}=2\mathrm{t}\end{array}$

Substituting in $\frac{\mathrm{a}}{\mathrm{l}}=\frac{-\mathrm{a}}{-\mathrm{t}}=\frac{4}{2\mathrm{t}}$,

$\begin{array}{rcl}\frac{\mathrm{a}}{\mathrm{l}}& =& \frac{-\mathrm{a}}{-\mathrm{t}}\\ \mathrm{l}& =& \mathrm{t}\end{array}$

Substituting $\mathrm{l}=\mathrm{t},\mathrm{m}=-\mathrm{t},\mathrm{n}=2\mathrm{t}\mathrm{in}\mathrm{lx}+\mathrm{my}+\mathrm{nz}=0$

$\begin{array}{rcl}\mathrm{tx}-\mathrm{ty}+2\mathrm{tz}& =& 0\\ \Rightarrow (\mathrm{x}-\mathrm{y}+2\mathrm{z})& =& 0\end{array}$

Step 4: Finding the length of BD

$\mathrm{BD}=\frac{(0-4+10)}{\sqrt{1^2+1^2+2^2}}\because \mathrm{coordinates}\mathrm{of}\mathrm{B}\mathrm{is}\mathrm{B}(0,4,5)$

$\begin{array}{rcl}\therefore \mathrm{BD}& =& \frac{(0-4+10)}{\sqrt{1+1+4}}\\ & =& \frac{6}{\sqrt{6}}\\ & =& \sqrt{6}\mathrm{units}\end{array}$

Step 5: Finding the distance of BC

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & \end{array}\mathrm{BC}=\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2+{({\mathrm{z}}_2-{\mathrm{z}}_1)}^2} \\ \begin{array}{rcl}& \Rightarrow & \end{array}\mathrm{BC}=\sqrt{0+{\left(6\right)}^2+{\left(6\right)}^2} \\ \begin{array}{rcl}& \Rightarrow & \end{array}\mathrm{BC}=\sqrt{72}\mathrm{units} \end{gathered}$$

Step 6: Finding the distance of CD

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & \end{array}\mathrm{CD}=\sqrt{{\mathrm{BC}}^2-{\mathrm{BD}}^2} \\ \begin{array}{rcl}& \Rightarrow & \end{array}\mathrm{CD}=\sqrt{72-6} \\ \begin{array}{rcl}& \Rightarrow & \end{array}\mathrm{CD}=\sqrt{66}\mathrm{units} \end{gathered}$$

Hence, the correct answer is an option (D).