find the point on the lines through the points A(1,2,3) and B(5,8,15) at a distance 14 units from the midpoint of AB

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{A}\left(1,2,3\right) \mathrm{and} \mathrm{B}\left(5,8,15\right) \\ \mathrm{Equation} \mathrm{of} \mathrm{line} \mathrm{AB} \mathrm{is} \\ \frac{\mathrm{x}-1}{5-1}=\frac{\mathrm{y}-2}{8-2}=\frac{\mathrm{z}-3}{15-3}=\mathrm{k} \\ \Rightarrow \frac{\mathrm{x}-1}{4}=\frac{\mathrm{y}-2}{6}=\frac{\mathrm{z}-3}{12}=\mathrm{k} \\ \mathrm{Hence} \mathrm{any} \mathrm{general} \mathrm{point} \mathrm{P}\hspace{0.17em}\mathrm{on} \mathrm{line} \mathrm{is} \mathrm{P}\left(4\mathrm{k}+1,6\mathrm{k}+2,12\mathrm{k}+3\right) \\ \mathrm{Mid} \mathrm{point} \mathrm{of} \mathrm{AB}\hspace{0.17em}\mathrm{is} \mathrm{C}\left(\frac{1+5}{2},\frac{2+8}{2},\frac{3+15}{2}\right)=\mathrm{C}\left(3,5,9\right) \\ \mathrm{Acording} \mathrm{to} \mathrm{question} \\ \mathrm{PC}=14 \\ \Rightarrow {\mathrm{PC}}^2=196 \\ \Rightarrow {\left(4\mathrm{k}+1-3\right)}^2+{\left(6\mathrm{k}+2-5\right)}^2+{\left(12\mathrm{k}+3-9\right)}^2=196 \\ \Rightarrow {\left(4\mathrm{k}-2\right)}^2+{\left(6\mathrm{k}-3\right)}^2+{\left(12\mathrm{k}-6\right)}^2=196 \\ \Rightarrow 16{\mathrm{k}}^2+4-16\mathrm{k}+36{\mathrm{k}}^2+9-36\mathrm{k}+144{\mathrm{k}}^2+36-144\mathrm{k}=196 \\ \Rightarrow 196{\mathrm{k}}^2-196\mathrm{k}-147=0 \\ \Rightarrow 4{\mathrm{k}}^2-4\mathrm{k}-3=0 \\ \Rightarrow 4{\mathrm{k}}^2-6\mathrm{k}+2\mathrm{k}-3=0 \\ \Rightarrow 2\mathrm{k}\left(2\mathrm{k}-3\right)+1\left(2\mathrm{k}-3\right)=0 \\ \Rightarrow \left(2\mathrm{k}-3\right)\left(2\mathrm{k}+1\right)=0 \\ \Rightarrow \mathrm{k}=\frac{3}{2} \mathrm{or} \mathrm{k}=-\frac{1}{2} \\ \mathrm{P}\left(4\mathrm{k}+1,6\mathrm{k}+2,12\mathrm{k}+3\right) \\ \mathrm{P}\left(4\mathrm{k}+1,6\mathrm{k}+2,12\mathrm{k}+3\right)=\left(4\times \frac{3}{2}+1,6\times \frac{3}{2}+2,12\times \frac{3}{2}+3\right)=\left(\mathbf{7}\mathbf{,}\mathbf{11}\mathbf{,}\mathbf{21}\right) \\ \mathrm{or} \\ \mathrm{P}\left(4\mathrm{k}+1,6\mathrm{k}+2,12\mathrm{k}+3\right)=\left(4\times \left(-\frac{1}{2}\right)+1,6\times \left(-\frac{1}{2}\right)+2,12\times \left(-\frac{1}{2}\right)+3\right)=\left(\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{3}\right) \end{gathered}$$

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