$\frac{x-3}{1}$ = $\frac{y-5}{-2}$ = $\frac{z-7}{1}$ The vector form of this equation is: $\vec{r}$ = $\hat{3i}+\hat{5j}+\hat{7k}$ + λ $(\hat{i}-\hat{2j}+\hat{k})$ $\vec{r}$ = $\overset{\rightarrow_1}{a} + \lambda\overset{\rightarrow}{b_1}$ ... (1) $\frac{x+1}{7}$ = $\frac{y+1}{-6}$ = $\frac{z+1}{1}$ The vector form of this equation is: $\vec{r}$ = - $\hat{i}-\hat{j}-\hat{k}$ + λ $(\hat{7i}-\hat{6j}+\hat{k})$ $\vec{r}$ = $\vec{a}_2+\lambda\vec{b}_2$ Therefore, $\vec{a}_1$ = $\hat{3i}+\hat{5j}+\hat{7k}$ , $\vec{b}_1$ = $\hat{i}-\hat{2j}+\hat{k}$ , $\vec{a}_2$ = - $\hat{i}-\hat{j}-\hat{k}$ and $\vec{b}_2$ = $\hat{7i}-\hat{6j}+\hat{k}$ Now, the shortest distance between these two lines is given by: d = $\left| \frac{\vec{b}_1 \times \vec{b}_2 \cdot \vec{a}_2 - \vec{a}_1}{|\vec{b}_1 \times \vec{b}_1|} \right|$ $\vec{b}_1 \times \vec{b}_2$ = $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 7 & -6 & 1 \end{vmatrix}$ = $\hat{i}(2+6)$ - $\hat{j}(1-7)$ + $\hat{k}(-6+14)$ = $\hat{4i}+\hat{6j}+\hat{8k}$ $|\vec{b}_1 \times \vec{b}_2|$ = $\sqrt{4^2+6^2+8^2}$ = $\sqrt{116}$ $\vec{a}_2 - \vec{a}_1$ = $-\hat{i}-\hat{j}-\hat{k}$ - $\hat{3i}+\hat{5j}+\hat{7k}$ = $-\hat{4i}-\hat{6j}-\hat{8k}$ ∴ d = = $\left| \frac{-16-36-64}{\sqrt{116}} \right|$ = $\left| \frac{-116}{\sqrt{116}} \right|$ = $\sqrt{116}$ OR Let $\frac{x+2}{3}$ = $\frac{y+1}{2}$ = $\frac{z-3}{2}$ = λ x = 2 + 3 λ ,y = - 1 + 2 λ ,z = 3 + 2 λ Therefore, a point on this line is: {(-2+3λ), (-1 + 2λ), (3 + 2λ)} The distance of the point{(-2+3λ), (-1 + 2λ), (3 + 2λ)} from point (1, 2, 3) = $3\sqrt{2}$ ∴ = $3\sqrt{2}$ ⇒ - 3 + $3\lambda^2$ + (-3) + $2\lambda^+2\lambda^2$ = 18 ⇒ 9 + $9\lambda^2$ - 18λ + 9 + $4\lambda^2$ - 12λ + $4\lambda^2$ = 18 $17\lambda^2$ - 30λ = 0 λ = 0 , λ = $\frac{30}{17}$ When λ = $\frac{30}{17}$ x = - 2 + 3λ = - 2 + 3 $\left(\frac{30}{17}\right)$ = - 2 + $\frac{90}{17}$ = $\frac{56}{17}$ y = - 1 + 2λ = - 1 + 2 $\left(\frac{30}{17}\right)$ = - 1 + $\frac{60}{17}$ = $\frac{43}{17}$ z = 3 + 2λ = 3 + 2 $\left(\frac{30}{17}\right)$ = $\frac{51+60}{17}$ = $\frac{111}{17}$ Thus, when λ = $\frac{30}{17}$ , the point is $\left(\frac{56}{17},\frac{43}{17},\frac{111}{17}\right)$ and when λ = 0 , the point is (- 2 , - 1 , 3)