The distance of the point $(-1,2,3)$ from the plane $\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10$ parallel to the line of the shortest distance between the lines $\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})$ and $\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})$ is :

1. Determine the line of shortest distance between the given two lines: <br/><br/>Direction vector of line 1: $\vec{d_1} = 2\hat{i} + \hat{k}$ <br/><br/>Direction vector of line 2: $\vec{d_2} = \hat{i} - \hat{j} + \hat{k}$ <br/><br/>Now, let's find the cross product $\vec{N} = \vec{d_1} \times \vec{d_2}$ <br/><br/>$$ \vec{N} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & -1 & 1 \\ \end{vmatrix} = (\hat{i}(0+1) - \hat{j}(2-1) + \hat{k}(-2-0)) = \hat{i} - \hat{j} - 2\hat{k} $$ <br/><br/>So, the direction vector of the line of shortest distance is $\vec{N} = \hat{i} - \hat{j} - 2\hat{k}$. <br/><br/>2. Find the equation of a line passing through point $A(-1, 2, 3)$ and having the direction vector $\vec{N}$: <br/><br/>$\frac{x + 1}{1} = \frac{y - 2}{-1} = \frac{z - 3}{-2} = \lambda$ <br/><br/>3. Find the point $P$ where the line intersects the given plane: <br/><br/>Let the coordinates of point $P$ be in terms of $\lambda$: <br/><br/>$P(\lambda - 1, -\lambda + 2, -2\lambda + 3)$ <br/><br/>Since $P$ lies on the plane $x - 2y + 3z = 10$, we can substitute the coordinates of $P$ in terms of $\lambda$ into the equation of the plane: <br/><br/>$(\lambda - 1) - 2(-\lambda + 2) + 3(-2\lambda + 3) = 10$ <br/><br/>$\lambda = -2$ <br/><br/>$P(-3, 4, 7)$ <br/><br/>4. Calculate the distance between points $A$ and $P$: <br/><br/>$AP = \sqrt{(-3 - (-1))^2 + (4 - 2)^2 + (7 - 3)^2} = 2\sqrt{6}$