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Let ' $\mathrm{O}$ ' be the origin, $\mathrm{A}$ and $\mathrm{B}$ be two points with position vectors $-3 \hat{i}-3 \hat{j}+4 \hat{k}$ and $4 \hat{i}-4 \hat{j}-3 \hat{k}$ respectively. Let $P$ be a point such that the line drawn through $P$ parallel to $\overrightarrow{\mathrm{OB}}$ meets $\mathrm{OA}$ in $\mathrm{L}$ and another line through $\mathrm{P}$ parallel to $\overrightarrow{\mathrm{OA}}$ meets $\mathrm{OB}$ in $\mathrm{M}$. If $\mathrm{L}$ divides $\mathrm{OA}$ in the ratio $2: 3$ and $\mathrm{M}$ divides $\mathrm{OB}$ in the ratio $3: 2$, then the distance from 0 to $\mathrm{P}$ is
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$\frac{19}{5}$
Position vector of $\mathrm{L}$ is $\frac{2}{5}(-3 \hat{i}-3 \hat{j}+4 \hat{k})$
Position vector of $\mathrm{M}$ is $=\frac{3}{5}(4 \hat{i}-4 \hat{j}+3 \hat{k})$
Since $\overrightarrow{\mathrm{LP}} \|$ and $\overrightarrow{\mathrm{OB}}$ and $\overrightarrow{\mathrm{PM}} \| \overrightarrow{\mathrm{OA}}$, therefore OMPL is parallelogram.
$\begin{aligned} & \because \overrightarrow{O P}=\overrightarrow{O L}+\overrightarrow{O M}=-(-6 \hat{i}-6 \hat{j}+8 \hat{k}+12 \hat{i}-12 \hat{j}-9 \hat{k}) \\ & =\frac{1}{5}(-6 \hat{i}-18 \hat{j}-\hat{k}) \\ & |\overrightarrow{\mathrm{OP}}|=\frac{1}{5} \sqrt{36+324+1}=\frac{19}{5}\end{aligned}$
Position vector of $\mathrm{M}$ is $=\frac{3}{5}(4 \hat{i}-4 \hat{j}+3 \hat{k})$
Since $\overrightarrow{\mathrm{LP}} \|$ and $\overrightarrow{\mathrm{OB}}$ and $\overrightarrow{\mathrm{PM}} \| \overrightarrow{\mathrm{OA}}$, therefore OMPL is parallelogram.
$\begin{aligned} & \because \overrightarrow{O P}=\overrightarrow{O L}+\overrightarrow{O M}=-(-6 \hat{i}-6 \hat{j}+8 \hat{k}+12 \hat{i}-12 \hat{j}-9 \hat{k}) \\ & =\frac{1}{5}(-6 \hat{i}-18 \hat{j}-\hat{k}) \\ & |\overrightarrow{\mathrm{OP}}|=\frac{1}{5} \sqrt{36+324+1}=\frac{19}{5}\end{aligned}$
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