If lines x-1/2=y+1/3=z-1/4 and x-3/1=y-k/2=z/1 intersect,then find the value of k and hence, find the equation of the plane containing these lines

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ {\mathrm{L}}_1:\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-1}{4}=\mathrm{u}, \mathrm{where} \text{'}\mathrm{u}\text{'} \mathrm{is} \mathrm{parameter}. \\ \mathrm{Direction} \mathrm{vector} \mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by}: \\ \overrightarrow{{\mathrm{l}}_1}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}} \\ \mathrm{Any} \mathrm{general} \mathrm{point} \mathrm{on} \mathrm{line} \mathrm{will} \mathrm{be}: \\ \left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(2\mathrm{u}+1,3\mathrm{u}-1,4\mathrm{u}+1\right) \\ {\mathrm{L}}_2:\frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}=\mathrm{v}, \mathrm{where} \text{'}\mathrm{v}\text{'} \mathrm{is} \mathrm{parameter}. \\ \mathrm{Direction} \mathrm{vector} \mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by}: \\ \overrightarrow{{\mathrm{l}}_2}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \mathrm{Any} \mathrm{general} \mathrm{point} \mathrm{on} \mathrm{line} \mathrm{will} \mathrm{be}: \\ \left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{v}+3,2\mathrm{v}+\mathrm{k},\mathrm{v}\right) \\ \mathrm{If} \mathrm{two} \mathrm{lines} \mathrm{intersect}, \mathrm{then} \mathrm{for} \mathrm{some} \text{'}\mathrm{u}\text{'} \mathrm{and} \text{'}\mathrm{v}\text{'}, \\ \left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(2\mathrm{u}+1,3\mathrm{u}-1,4\mathrm{u}+1\right)=\left(\mathrm{v}+3,2\mathrm{v}+\mathrm{k},\mathrm{v}\right) \\ \Rightarrow 2\mathrm{u}+1=\mathrm{v}+3 \\ \Rightarrow 2\mathrm{u}-\mathrm{v}=2 ;\left(\mathrm{i}\right) \\ 4\mathrm{u}+1=\mathrm{v} \\ \Rightarrow 4\mathrm{u}-\mathrm{v}=-1 ;\left(\mathrm{ii}\right) \\ \left(\mathrm{ii}\right)-\left(\mathrm{i}\right) \\ 4\mathrm{u}-\mathrm{v}-2\mathrm{u}+\mathrm{v}=-1-2 \\ 2\mathrm{u}=-3 \\ \Rightarrow \mathrm{u}=-\frac{3}{2} \\ \mathrm{Put} \mathrm{this} \mathrm{in} \left(\mathrm{i}\right) \\ -3-\mathrm{v}=2 \\ \Rightarrow \mathrm{v}=-5 \\ \mathrm{y} \mathrm{co}-\mathrm{ordinate} \mathrm{will} \mathrm{also} \mathrm{be} \mathrm{same}. \\ 3\mathrm{u}-1=2\mathrm{v}+\mathrm{k} \\ 3\left(-\frac{3}{2}\right)-1=2\left(-5\right)+\mathrm{k} \\ \Rightarrow -\frac{9}{2}-1+10=\mathrm{k} \\ \Rightarrow \mathrm{k}=-\frac{9}{2}+9=\frac{-9+18}{2} \\ \Rightarrow \mathbf{k}\mathbf{=}\frac{\mathbf{9}}{\mathbf{2}}\mathbf{ }\left(\mathbf{Answer}\right) \\ \mathrm{Point} \mathrm{of} \mathrm{intersection}=\mathrm{P}\left(2\mathrm{u}+1,3\mathrm{u}-1,4\mathrm{u}+1\right) \\ =\mathrm{P}\left(-3+1,-\frac{9}{2}-1,-6+1\right)=\mathbf{P}\left(\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{-}\frac{\mathbf{11}}{\mathbf{2}}\mathbf{,}\mathbf{-}\mathbf{5}\right) \\ \mathrm{Let} \mathrm{required} \mathrm{equation} \mathrm{of} \mathrm{plane} \mathrm{be} \mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}=0 \\ \mathrm{Now} \mathrm{direction} \mathrm{normal} \mathrm{of} \mathrm{plane} \mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by} \mathrm{cross} \mathrm{product} \mathrm{of} \overrightarrow{{\mathrm{l}}_1} \mathrm{and} \overrightarrow{{\mathrm{l}}_2}. \\ \Rightarrow \mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}+\mathrm{c}\hat{\mathrm{k}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 3& 4\\ 1& 2& 1\end{array}\right| \\ =\hat{\mathrm{i}}\left(3-8\right)-\left(2-4\right)\hat{\mathrm{j}}+\left(4-3\right)\hat{\mathrm{k}} \\ \Rightarrow \mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}+\mathrm{c}\hat{\mathrm{k}}=-5\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ \Rightarrow \mathrm{a}=-5 \mathrm{and} \mathrm{b}=2 \mathrm{nad} \mathrm{c}=1 \\ \mathrm{Hence} \mathrm{plane} \mathrm{i} \\ -5\mathrm{x}+2\mathrm{y}+\mathrm{z}+\mathrm{d}=0 \\ \mathrm{As} \mathrm{plane} \mathrm{contains} {\mathrm{L}}_1:\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-1}{4}, \mathrm{hence} \\ \mathrm{it} \mathrm{will} \mathrm{contain} \mathrm{point} \left(1,-1,1\right) \\ \Rightarrow -5-2+1+\mathrm{d}=0 \\ \Rightarrow \mathrm{d}=6 \\ \Rightarrow \mathbf{-}\mathbf{5}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{z}\mathbf{+}\mathbf{6}\mathbf{=}\mathbf{0}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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