Find the coordinates of the foot of the perpendicular drawn from the point (1, 2, 3) to the line
Find the coordinates of the foot of the perpendicular drawn from the point (1, 2, 3) to the line

Find the coordinates of the foot of the perpendicular drawn from the point to the line . Also, find the length of the perpendicular from the given point to the line.
Given: Equation of line is .
To find: coordinates of foot of the perpendicular from to the line. And find the length of the perpendicular.
Formula Used:
1. Equation of a line is
Cartesian form:
where is a point on the line and is the direction ratios of the line.
2. Distance between two points and is

$\sqrt{{\left({\mathrm{x}}_{1}–{\mathrm{x}}_{2}\right)}^{2}+{\left({\mathrm{y}}_{1}–{\mathrm{y}}_{2}\right)}^{2}+{\left({\mathrm{z}}_{1}–{\mathrm{z}}_{2}\right)}^{2}}$
\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}+\left(\mathrm{z}_{1}-\mathrm{z}_{2}\right)^{2}}

Explanation:
Let

$\frac{x–6}{\mathrm{\not}3}=\frac{y–7}{3}=\frac{z–7}{–3}=\lambda$
\frac{x-6}{\not 3}=\frac{y-7}{3}=\frac{z-7}{-3}=\lambda

So the foot of the perpendicular is
Direction ratio of the line is
Direction ratio of the perpendicular is

$\begin{array}{l}⇒\left(3\lambda +6–1\right):\left(2\lambda +7–2\right):\left(–2\lambda +7–3\right)\\ ⇒\left(3\lambda +5\right):\left(2\lambda +5\right):\left(–2\lambda +4\right)\end{array}$
\begin{array}{l}
\Rightarrow(3 \lambda+6-1):(2 \lambda+7-2):(-2 \lambda+7-3) \\
\Rightarrow(3 \lambda+5):(2 \lambda+5):(-2 \lambda+4)
\end{array}

Since this is perpendicular to the line,

$\begin{array}{l}3\left(3\lambda +5\right)+2\left(2\lambda +5\right)–2\left(–2\lambda +4\right)=0\\ ⇒9\lambda +15+4\lambda +10+4\lambda –8=0\\ ⇒17\lambda =–17\\ ⇒\lambda =–1\end{array}$
\begin{array}{l}
3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4)=0 \\
\Rightarrow 9 \lambda+15+4 \lambda+10+4 \lambda-8=0 \\
\Rightarrow 17 \lambda=-17 \\
\Rightarrow \lambda=-1
\end{array}

So the foot of the perpendicular is

\begin{array}{l}
\text { Distance }=\sqrt{(3-1)^{2}+(5-2)^{2}+(9-3)^{2}} \\
=\sqrt{4+9+36} \\
=7 \text { units }
\end{array}