Find the equation of the plane through the line :
`(x - 1)/(3) = (y - 4)/(2) = (z - 4)/(-2)`
and parallel to the line :
`(x + 1)/(2) = (y-1)/(4) = (z + 2)/(1)`.
Hence, find the shortest distance between the lines.

Answer

To solve the problem, we need to find the equation of a plane that contains the line given by:

\[ \frac{x - 1}{3} = \frac{y - 4}{2} = \frac{z - 4}{-2} \]

and is parallel to the line given by:

\[ \frac{x + 1}{2} = \frac{y - 1}{4} = \frac{z + 2}{1} \]

Then, we will find the shortest distance between the two lines.

### Step 1: Identify the direction ratios of the lines

For the first line (L1), the direction ratios can be extracted from the coefficients of x, y, and z:

- Direction ratios of L1: \( \vec{d_1} = (3, 2, -2) \)

For the second line (L2), we have:

- Direction ratios of L2: \( \vec{d_2} = (2, 4, 1) \)

### Step 2: Find the normal vector of the plane

The normal vector \( \vec{n} \) of the plane can be found using the cross product of the direction ratios of the two lines:

\[ \vec{n} = \vec{d_1} \times \vec{d_2} = (3, 2, -2) \times (2, 4, 1) \]

Calculating the cross product:

\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -2 \\ 2 & 4 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{n} = \hat{i}(2 \cdot 1 - (-2) \cdot 4) - \hat{j}(3 \cdot 1 - (-2) \cdot 2) + \hat{k}(3 \cdot 4 - 2 \cdot 2) \]

\[ = \hat{i}(2 + 8) - \hat{j}(3 + 4) + \hat{k}(12 - 4) \]

\[ = 10\hat{i} - 7\hat{j} + 8\hat{k} \]

Thus, the normal vector is:

\[ \vec{n} = (10, -7, 8) \]

### Step 3: Use a point on the line to find the equation of the plane

The plane contains the point \( A(1, 4, 4) \) from line L1. The equation of the plane can be expressed as:

\[ 10(x - 1) - 7(y - 4) + 8(z - 4) = 0 \]

Expanding this:

\[ 10x - 10 - 7y + 28 + 8z - 32 = 0 \]

Combining like terms:

\[ 10x - 7y + 8z - 14 = 0 \]

Thus, the equation of the plane is:

\[ 10x - 7y + 8z = 14 \]

### Step 4: Find the shortest distance between the two lines

To find the shortest distance \( d \) between the two lines, we can use the formula:

\[ d = \frac{|(\vec{a_1} - \vec{a_2}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} \]

Where \( \vec{a_1} = (1, 4, 4) \) (a point on L1) and \( \vec{a_2} = (-1, 1, -2) \) (a point on L2).

Calculating \( \vec{a_1} - \vec{a_2} \):

\[ \vec{a_1} - \vec{a_2} = (1 - (-1), 4 - 1, 4 - (-2)) = (2, 3, 6) \]

Now we need to calculate \( \vec{d_1} \times \vec{d_2} \):

We already found \( \vec{n} = (10, -7, 8) \), so:

\[ |\vec{d_1} \times \vec{d_2}| = \sqrt{10^2 + (-7)^2 + 8^2} = \sqrt{100 + 49 + 64} = \sqrt{213} \]

Now calculate the dot product \( (\vec{a_1} - \vec{a_2}) \cdot \vec{n} \):

\[ (2, 3, 6) \cdot (10, -7, 8) = 2 \cdot 10 + 3 \cdot (-7) + 6 \cdot 8 = 20 - 21 + 48 = 47 \]

Finally, substituting into the distance formula:

\[ d = \frac{|47|}{\sqrt{213}} = \frac{47}{\sqrt{213}} \]

### Final Answer

The equation of the plane is:

\[ 10x - 7y + 8z = 14 \]

The shortest distance between the lines is:

\[ d = \frac{47}{\sqrt{213}} \]