Consider the lines `x=y=z` and line `2x+y+z-1=0=3x+y+2z-2`, then

Options

A. The shortest distance between the two lines is `(1)/(sqrt(2))`
B. The shortest distance between the two lines is `sqrt(2)`
C. Plane containing the line `L_(2)` and parallel to line `L_(1)` is `z-x+1=0`
D. Perpendicular distance of origin from plane containing line `L_(2)` and parallel to line `L_(1)` is `(1)/(sqrt(2))`

Answer

To solve the problem, we need to find the shortest distance between the line defined by \( x = y = z \) and the line formed by the intersection of the two planes given by the equations \( 2x + y + z - 1 = 0 \) and \( 3x + y + 2z - 2 = 0 \).

### Step 1: Identify the direction vector of the line \( x = y = z \)

The line \( x = y = z \) can be represented in vector form as:
\[ \mathbf{r_1} = t(1, 1, 1) \]
where \( t \) is a parameter. The direction vector \( \mathbf{b_1} \) of this line is:
\[ \mathbf{b_1} = (1, 1, 1) \]

### Step 2: Find the normal vectors of the planes

For the first plane \( 2x + y + z - 1 = 0 \), the normal vector \( \mathbf{n_1} \) is:
\[ \mathbf{n_1} = (2, 1, 1) \]

For the second plane \( 3x + y + 2z - 2 = 0 \), the normal vector \( \mathbf{n_2} \) is:
\[ \mathbf{n_2} = (3, 1, 2) \]

### Step 3: Find the direction vector of the line of intersection of the two planes

The direction vector \( \mathbf{b_2} \) of the line formed by the intersection of the two planes can be found by taking the cross product of the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \):
\[ \mathbf{b_2} = \mathbf{n_1} \times \mathbf{n_2} \]

Calculating the cross product:
\[ \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 3 & 1 & 2 \end{vmatrix} = \mathbf{i}(1 \cdot 2 - 1 \cdot 1) - \mathbf{j}(2 \cdot 2 - 1 \cdot 3) + \mathbf{k}(2 \cdot 1 - 1 \cdot 3) \]
\[ = \mathbf{i}(2 - 1) - \mathbf{j}(4 - 3) + \mathbf{k}(2 - 3) \]
\[ = \mathbf{i}(1) - \mathbf{j}(1) - \mathbf{k}(1) = (1, -1, -1) \]

### Step 4: Find a point on the line of intersection

To find a point on the line of intersection of the two planes, we can set \( z = 0 \) and solve the equations:
1. \( 2x + y - 1 = 0 \)
2. \( 3x + y - 2 = 0 \)

From the first equation, we have:
\[ y = 1 - 2x \]
Substituting into the second equation:
\[ 3x + (1 - 2x) - 2 = 0 \implies 3x - 2x + 1 - 2 = 0 \implies x - 1 = 0 \implies x = 1 \]
Then substituting \( x = 1 \) back into \( y = 1 - 2x \):
\[ y = 1 - 2(1) = -1 \]
So, a point on the line of intersection is \( (1, -1, 0) \).

### Step 5: Use the formula for the shortest distance between two skew lines

The formula for the shortest distance \( d \) between two skew lines is:
\[ d = \frac{|(\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]
Where:
- \( \mathbf{a_1} = (0, 0, 0) \) (a point on the line \( x = y = z \))
- \( \mathbf{a_2} = (1, -1, 0) \) (a point on the line of intersection)

### Step 6: Calculate \( \mathbf{a_2} - \mathbf{a_1} \)

\[ \mathbf{a_2} - \mathbf{a_1} = (1, -1, 0) - (0, 0, 0) = (1, -1, 0) \]

### Step 7: Calculate \( \mathbf{b_1} \times \mathbf{b_2} \)

\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 1 & -1 & -1 \end{vmatrix} = \mathbf{i}(1 \cdot -1 - 1 \cdot -1) - \mathbf{j}(1 \cdot -1 - 1 \cdot 1) + \mathbf{k}(1 \cdot -1 - 1 \cdot 1) \]
\[ = \mathbf{i}(-1 + 1) - \mathbf{j}(-1 - 1) + \mathbf{k}(-1 - 1) \]
\[ = \mathbf{i}(0) + \mathbf{j}(2) + \mathbf{k}(-2) = (0, 2, -2) \]

### Step 8: Calculate the magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \)

\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{0^2 + 2^2 + (-2)^2} = \sqrt{0 + 4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Step 9: Calculate the dot product \( (\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1}) \)

\[ (\mathbf{b_1} \times \mathbf{b_2}) \cdot (1, -1, 0) = (0, 2, -2) \cdot (1, -1, 0) = 0 \cdot 1 + 2 \cdot -1 + -2 \cdot 0 = -2 \]

### Step 10: Substitute into the distance formula

\[ d = \frac{| -2 |}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer

The shortest distance between the two lines is:
\[ \boxed{\frac{1}{\sqrt{2}}} \]