Part 1: Finding the vector and Cartesian equations of the line and the distance between the two lines.

Step 1: Find the direction vector of the line joining A(3,3,-5) and B(1,0,-11).

Direction vector, \( \vec{d} = \vec{B} - \vec{A} = (1-3)\hat{i} + (0-3)\hat{j} + (-11-(-5))\hat{k} = -2\hat{i} - 3\hat{j} - 6\hat{k} \)

Step 2: Write the vector equation of the line passing through (1,2,-4) and parallel to \( \vec{d} \).

\( \vec{r} = (1\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(-2\hat{i} - 3\hat{j} - 6\hat{k}) \), where \( \lambda \) is a scalar.

Step 3: Write the Cartesian equation of the line.

\( \frac{x-1}{-2} = \frac{y-2}{-3} = \frac{z+4}{-6} \)

Step 4: Find the vector equation of the line joining A(3,3,-5) and B(1,0,-11).

\( \vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(-2\hat{i} - 3\hat{j} - 6\hat{k}) \), where \( \mu \) is a scalar.

Step 5: Find the shortest distance between the two parallel lines.

Let \( \vec{a_1} = \hat{i} + 2\hat{j} - 4\hat{k} \) and \( \vec{a_2} = 3\hat{i} + 3\hat{j} - 5\hat{k} \). Then \( \vec{a_2} - \vec{a_1} = 2\hat{i} + \hat{j} - \hat{k} \).

The distance \( d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{d}|}{|\vec{d}|} \).

\( (\vec{a_2} - \vec{a_1}) \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ -2 & -3 & -6 \end{vmatrix} = (-6-3)\hat{i} - (-12-2)\hat{j} + (-6+2)\hat{k} = -9\hat{i} + 14\hat{j} - 4\hat{k} \)

\( |(\vec{a_2} - \vec{a_1}) \times \vec{d}| = \sqrt{(-9)^2 + (14)^2 + (-4)^2} = \sqrt{81 + 196 + 16} = \sqrt{293} \)

\( |\vec{d}| = \sqrt{(-2)^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \)

\( d = \frac{\sqrt{293}}{7} \)

Part 2: OR Finding the equations of the line and the coordinates of the points.

Step 1: Find the direction vector of the line joining A(1,2,3) and B(3,5,9).

Direction vector, \( \vec{d} = \vec{B} - \vec{A} = (3-1)\hat{i} + (5-2)\hat{j} + (9-3)\hat{k} = 2\hat{i} + 3\hat{j} + 6\hat{k} \)

Step 2: Write the vector equation of the line passing through A(1,2,3) and B(3,5,9).

\( \vec{r} = (1\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \), where \( \lambda \) is a scalar.

Step 3: Write the Cartesian equation of the line.

\( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{6} = \lambda \)

Step 4: Find the coordinates of a general point on the line.

\( x = 2\lambda + 1, y = 3\lambda + 2, z = 6\lambda + 3 \). So the general point is \( (2\lambda + 1, 3\lambda + 2, 6\lambda + 3) \).

Step 5: Find the distance between the general point and B(3,5,9).

\( \sqrt{(2\lambda + 1 - 3)^2 + (3\lambda + 2 - 5)^2 + (6\lambda + 3 - 9)^2} = 14 \)

\( \sqrt{(2\lambda - 2)^2 + (3\lambda - 3)^2 + (6\lambda - 6)^2} = 14 \)

\( (2\lambda - 2)^2 + (3\lambda - 3)^2 + (6\lambda - 6)^2 = 196 \)

\( 4(\lambda - 1)^2 + 9(\lambda - 1)^2 + 36(\lambda - 1)^2 = 196 \)

\( 49(\lambda - 1)^2 = 196 \)

\( (\lambda - 1)^2 = 4 \)

\( \lambda - 1 = \pm 2 \)

\( \lambda = 3 \) or \( \lambda = -1 \)

Step 6: Find the coordinates of the points.

For \( \lambda = 3 \): \( x = 2(3) + 1 = 7, y = 3(3) + 2 = 11, z = 6(3) + 3 = 21 \). Point is (7, 11, 21).

For \( \lambda = -1 \): \( x = 2(-1) + 1 = -1, y = 3(-1) + 2 = -1, z = 6(-1) + 3 = -3 \). Point is (-1, -1, -3).