The line `l_(1)` passes through the point `(2,6,2)` and is perpendicular to the plane `2x+y-2z=10`. Then the shortest distance between the line `l_(1)` and the line `(x+1)/(2)=(y+4)/(-3)=(z)/(2)` is :

Options

A. `(19)/(3)`
B. `7`
C. `9`
D. `(13)/(3)`

Answer

To find the shortest distance between the two lines given in the problem, we will follow these steps:

### Step 1: Identify the Direction Ratios and Points of the Lines
The first line \( l_1 \) passes through the point \( (2, 6, 2) \) and is perpendicular to the plane \( 2x + y - 2z = 10 \). The normal vector of the plane, which gives the direction of line \( l_1 \), is \( \vec{n} = (2, 1, -2) \).

The second line \( l_2 \) is given in symmetric form:
\[ \frac{x + 1}{2} = \frac{y + 4}{-3} = \frac{z}{2} \]
From this, we can extract:
- A point on line \( l_2 \): \( (-1, -4, 0) \)
- The direction ratios of line \( l_2 \): \( (2, -3, 2) \)

### Step 2: Write the Vector Form of the Lines
The vector form of line \( l_1 \) can be written as:
\[ \vec{r_1} = (2, 6, 2) + t(2, 1, -2) \]
The vector form of line \( l_2 \) can be written as:
\[ \vec{r_2} = (-1, -4, 0) + s(2, -3, 2) \]

### Step 3: Find the Vector Connecting Points on Both Lines
Let \( A = (2, 6, 2) \) be a point on line \( l_1 \) and \( B = (-1, -4, 0) \) be a point on line \( l_2 \). The vector \( \vec{AB} \) is given by:
\[ \vec{AB} = B - A = (-1 - 2, -4 - 6, 0 - 2) = (-3, -10, -2) \]

### Step 4: Find the Direction Vectors
Let \( \vec{m} = (2, 1, -2) \) be the direction vector of line \( l_1 \) and \( \vec{n} = (2, -3, 2) \) be the direction vector of line \( l_2 \).

### Step 5: Calculate the Cross Product of the Direction Vectors
The cross product \( \vec{m} \times \vec{n} \) is computed as follows:
\[ \vec{m} \times \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i}(1 \cdot 2 - (-2)(-3)) - \hat{j}(2 \cdot 2 - (-2)(2)) + \hat{k}(2 \cdot (-3) - 1 \cdot 2) \]
\[ = \hat{i}(2 - 6) - \hat{j}(4 - 4) + \hat{k}(-6 - 2) \]
\[ = \hat{i}(-4) + \hat{j}(0) + \hat{k}(-8) = (-4, 0, -8) \]

### Step 6: Calculate the Magnitude of the Cross Product
The magnitude of the cross product \( |\vec{m} \times \vec{n}| \) is:
\[ |\vec{m} \times \vec{n}| = \sqrt{(-4)^2 + 0^2 + (-8)^2} = \sqrt{16 + 0 + 64} = \sqrt{80} = 4\sqrt{5} \]

### Step 7: Calculate the Shortest Distance
The shortest distance \( d \) between the two lines is given by the formula:
\[ d = \frac{|\vec{AB} \cdot (\vec{m} \times \vec{n})|}{|\vec{m} \times \vec{n}|} \]
Calculating \( \vec{AB} \cdot (\vec{m} \times \vec{n}) \):
\[ \vec{AB} \cdot (-4, 0, -8) = (-3)(-4) + (-10)(0) + (-2)(-8) = 12 + 0 + 16 = 28 \]
Thus, the shortest distance is:
\[ d = \frac{|28|}{4\sqrt{5}} = \frac{28}{4\sqrt{5}} = \frac{7}{\sqrt{5}} = \frac{7\sqrt{5}}{5} \]

### Final Answer
The shortest distance between the lines \( l_1 \) and \( l_2 \) is \( \frac{7\sqrt{5}}{5} \).