• # question_answer Find the coordinates of point on line $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{6},$ which are at a distance of 3 units from the point  $(1,\,\,-2,\,\,3).$ OR Show that the lines $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda (3\hat{i}-\hat{j})$ and $\vec{r}-=(4\hat{i}-\hat{k})+\mu (2\hat{i}+3\hat{k})$ are coplanar. Also, find the equation of the plane containing them.

 Let $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{6}=\lambda$ Then, arbitrary point on the line is $(2\lambda +1,\,\,3\lambda -2,\,\,6\lambda +3)$ Suppose $P\,(2\lambda +1,\,\,3\lambda -2,\,\,6\lambda +3)$ be the point at a distance 3 units from the point $Q\,(1,\,\,-2,\,\,3)$ $\therefore$                              $PQ=3$ ${{(PQ)}^{2}}={{(3)}^{2}}=9$ $\Rightarrow$${{(2\lambda +1-1)}^{2}}+{{(3\lambda -2+2)}^{2}}+{{(6\lambda +3-3)}^{2}}=9$ $\Rightarrow$   ${{(2\lambda )}^{2}}+{{(3\lambda )}^{2}}+{{(6\lambda )}^{2}}=9\Rightarrow 49{{\lambda }^{2}}=9$ $\Rightarrow$               ${{\lambda }^{2}}=\frac{9}{49}$ $\Rightarrow$               $\lambda =\pm \frac{3}{7}$ $\therefore$Points are $\left( 2\left( \frac{3}{7} \right)+1,\,\,3\left( \frac{3}{7} \right)-2,\,\,6\left( \frac{3}{7} \right)+3 \right)$ $\left( 2\left( \frac{-\,3}{7} \right)+1,\,\,3\left( \frac{-\,3}{7} \right)-2,\,\,6\left( \frac{-\,3}{7} \right)+3 \right)$ $=\left( \frac{13}{7},\,\,\frac{-\,5}{7},\,\,\frac{39}{7} \right)$ or $\left( \frac{1}{7},\,\,\frac{-\,23}{7},\,\,\frac{3}{7} \right)$ OR Given lines are $\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda \,(3\,\hat{i}-\hat{j})$ and                   $\overrightarrow{r}=(4\,\hat{i}-\hat{k})+\mu \,(2\,\hat{i}+3\hat{k})$ On comparing both equations of lines with $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$respectively, we get $\overrightarrow{{{a}_{1}}}=\hat{i}+\hat{j}-\hat{k},\,\,\overrightarrow{{{b}_{1}}}=3\hat{i}-\hat{j}$ And      $\overrightarrow{{{a}_{2}}}=4\,\hat{i}-\hat{k},\,\,\overrightarrow{{{b}_{2}}}=2\hat{i}+3\hat{k}$ $\therefore$ $=\hat{i}\,(-\,3-0)-\hat{j}\,(9-0)+\hat{k}\,(0+2)$ $=-\,3\,\hat{i}-9\hat{j}+2\hat{k}$ and ${{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}}=(4\hat{i}-\hat{k})-(\hat{i}+\hat{j}-\hat{k})=3\hat{i}-\hat{j}$ Now, $({{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}})\cdot ({{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}})=(3\,\hat{i}-\hat{j})\cdot (-\,3\,\hat{i}-9\hat{j}+2\hat{k})$ $=-\,9+9=0$ Hence, the given lines are coplanar. Now, Cartesian equations of given lines are $\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$and $\frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}$. Then, equation of plane containing them is \Rightarrow \left| \begin{matrix}    x-1 & y-1 & z-1  \\    3 & -1 & 0  \\    2 & 0 & 3  \\ \end{matrix} \right|=0\]$\Rightarrow$$(x-1)(-3-0)-(y-1)(9-0)+(z+1)(0+2)=0$ $\Rightarrow$               $-\,3x+3-9y+9+2z+2=0$ $\therefore$                              $3x+9y-2z=14$