If the shortest distance between the lines (x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x)/(1)=(y)/(…

Question

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$, then the sum of all possible values of $\alpha$ is

Accepted Answer

$\begin{gathered} d=|\frac{(\vec{a}-\vec{b})\cdot {\vec{p}}_{1}	imes {\vec{p}}_{2}}{|{\vec{p}}_{1}	imes {\vec{p}}_{2}|}|=\frac{5}{\sqrt{6}} \ \vec{a}-\vec{b}=\hat{i}+2\hat{j}-2\hat{k} \ \overset{	o }{{p}_{1}}	imes \overset{	o }{{p}_{2}}=|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 2 \quad  3 \quad  4 \ 1 \quad  \alpha  \quad  1\end{array}| \ \, =\hat{i}(3-4\alpha )-\hat{j}(-2)+\hat{k}(2\alpha -3) \ (\vec{a}-\vec{b})\cdot |\overset{	o }{{p}_{1}}	imes \overset{	o }{{p}_{2}}|=3-4\alpha +4-4\alpha +6 \ =13-8\alpha \end{gathered}$

$\begin{gathered} |\frac{13-8\alpha }{\sqrt{(3-4\alpha {)}^{2}+4+(2\alpha -3{)}^{2}}}|=\frac{5}{\sqrt{6}} \ |\frac{13-8\alpha }{\sqrt{20{\alpha }^{2}-36\alpha +22}}|=\frac{5}{\sqrt{6}} \ =6(13-8\alpha {)}^{2}=25(20{\alpha }^{2}-36\alpha +22) \ =116{\alpha }^{2}+348\alpha -464=0 \ =-3 \end{gathered}$

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