Find the image of the point $(0,2,3)$ in the line $\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}$.
Answer
Given: Equation of line is $\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}$.
To find: image of point $(0,2,3)$
Formula Used: Equation of a line is
Vector form: $\overrightarrow{\mathrm{I}}=\overrightarrow{\mathrm{a}}+\mathrm{\lambda} \overrightarrow{\mathrm{b}}$
Cartesian form: $\frac{\mathrm{x}-\mathrm{r}_{1}}{\mathrm{~b}_{\mathrm{n}}}=\frac{\mathrm{y}-\mathrm{y}_{1}}{\mathrm{~b}_{\mathrm{z}}}=\frac{\mathrm{z}-\mathrm{z}_{1}}{\mathrm{k}_{\mathrm{s}}}=\lambda$
where $\vec{a}=x_{1} \hat{\imath}+y_{1} \hat{l}+z_{1} \hat{k}$ is a point on the line and $\vec{b}=b_{1} \hat{i}+b_{2} \hat{l}+b_{3} \hat{k}$ with $b_{1}: b_{2}: b_{3}$ being the direction ratios of the line.
If 2 lines of direction ratios $a_{1}: a_{2}: a_{3}$ and $b_{1}: b_{2}: b_{3}$ are perpendicular, then $a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}=0$ Mid-point of line segment joining $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ is

$\left(\frac{x_1–x_2}{2},\frac{y_1–y_2}{2},\frac{z_1–z_2}{2}\right)$
\left(\frac{x_{1}-x_{2}}{2}, \frac{y_{1}-y_{2}}{2}, \frac{z_{1}-z_{2}}{2}\right)

Explanation:
Let

$\frac{x+3}{5}=\frac{y–1}{2}=\frac{z+4}{3}=\lambda$
\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda

So the foot of the perpendicular is $(5 \lambda-3,2 \lambda+1,3 \lambda-4)$
The direction ratios of the perpendicular is

$\begin{array}{l}(5\lambda –3–0):(2\lambda +1–2):(3\lambda –4–3)\\ \Rightarrow (5\lambda –3):(2\lambda –1):(3\lambda –7)\end{array}$
\begin{array}{l}
(5 \lambda-3-0):(2 \lambda+1-2):(3 \lambda-4-3) \\
\Rightarrow(5 \lambda-3):(2 \lambda-1):(3 \lambda-7)
\end{array}

Direction ratio of the line is $5: 2: 3$