Find the image of the point (0, 2, 3) in the line

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Find the image of the point (0, 2, 3) in the line

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