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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{are} \mathrm{looking} \mathrm{for} \mathrm{perpendicular} \mathrm{distance}. \\ \mathrm{We} \mathrm{have}: \\ \overrightarrow{\mathrm{P}}=-2\hat{\mathrm{i}}+4\hat{\mathrm{j}}-5\hat{\mathrm{k}} \\ \mathrm{L}:\frac{\mathrm{x}+3}{3}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}+8}{6}=\mathrm{t} \\ \mathrm{Direction} \mathrm{vector} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \overrightarrow{\mathrm{DR}}=3\hat{\mathrm{i}}+5\hat{\mathrm{j}}+6\hat{\mathrm{k}} \\ \mathrm{Any} \mathrm{general} \mathrm{point} \mathrm{on} \mathrm{line} \mathrm{will} \mathrm{be}: \\ \overrightarrow{\mathrm{Q}}=\left(3\mathrm{t}-3\right)\hat{\mathrm{i}}+\left(5\mathrm{t}+4\right)\hat{\mathrm{j}}+\left(6\mathrm{t}-8\right)\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{PQ}}=\left[\left(3\mathrm{t}-3\right)\hat{\mathrm{i}}+\left(5\mathrm{t}+4\right)\hat{\mathrm{j}}+\left(6\mathrm{t}-8\right)\hat{\mathrm{k}}\right]-\left[-2\hat{\mathrm{i}}+4\hat{\mathrm{j}}-5\hat{\mathrm{k}}\right] \\ =\left(3\mathrm{t}-3+2\right)\hat{\mathrm{i}}+\left(5\mathrm{t}+4-4\right)\hat{\mathrm{j}}+\left(6\mathrm{t}-8+5\right)\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{PQ}}=\left(3\mathrm{t}-1\right)\hat{\mathrm{i}}+\left(5\mathrm{t}\right)\hat{\mathrm{j}}+\left(6\mathrm{t}-3\right)\hat{\mathrm{k}} \\ \overrightarrow{\mathrm{PQ}} \mathrm{will} \mathrm{be} \mathrm{perpendicular} \mathrm{to} \overrightarrow{\mathrm{DR}}, \mathrm{hence} \\ \overrightarrow{\mathrm{PQ}}.\overrightarrow{\mathrm{DR}}=0 \\ \Rightarrow \left[\left(3\mathrm{t}-1\right)\hat{\mathrm{i}}+\left(5\mathrm{t}\right)\hat{\mathrm{j}}+\left(6\mathrm{t}-3\right)\hat{\mathrm{k}}\right].\left[3\hat{\mathrm{i}}+5\hat{\mathrm{j}}+6\hat{\mathrm{k}}\right]=0 \\ \Rightarrow 3\left(3\mathrm{t}-1\right)+25\mathrm{t}+6\left(6\mathrm{t}-3\right)=0 \\ \Rightarrow 9\mathrm{t}-3+25\mathrm{t}+36\mathrm{t}-18=0 \\ \Rightarrow 70\mathrm{t}=21 \\ \Rightarrow \mathrm{t}=\frac{21}{70} \\ \mathrm{Hence} \mathrm{point} \mathrm{Q}\left(3\mathrm{t}-3, 5\mathrm{t}+4, 6\mathrm{t}-8\right)=\left(\frac{63}{70}-3,\frac{105}{70}+4, \frac{126}{70}-8\right) \\ =\left(-\frac{147}{70},\frac{385}{70}, -\frac{434}{70}\right) \\ \mathrm{Now} \mathrm{required} \mathrm{distance} \mathrm{can} \mathrm{be} \mathrm{found} \mathrm{by} \mathrm{distance} \mathrm{formula}. \\ \mathrm{PQ}=\sqrt{{\left(-\frac{147}{70}+2\right)}^2+{\left(\frac{385}{70}-4\right)}^2+{\left(-\frac{434}{70}+5\right)}^2} \\ =\sqrt{{\left(-\frac{7}{70}\right)}^2+{\left(\frac{105}{70}\right)}^2+{\left(-\frac{84}{70}\right)}^2} \\ =\frac{\sqrt{49+11025+7056}}{70} \\ =\frac{\sqrt{18130}}{70} \end{gathered}$$
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