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 \( x = y = z \) and 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 planes and the line
The first plane \( P_1 \) is given by the equation:
\[ 2x + y + z - 1 = 0 \]
The second plane \( P_2 \) is given by the equation:
\[ 3x + y + 2z - 2 = 0 \]
The line is defined by:
\[ x = y = z \]

### Step 2: Find the normal vectors of the planes
The normal vector \( \mathbf{n_1} \) of the first plane \( P_1 \) can be derived from its coefficients:
\[ \mathbf{n_1} = \langle 2, 1, 1 \rangle \]
The normal vector \( \mathbf{n_2} \) of the second plane \( P_2 \) is:
\[ \mathbf{n_2} = \langle 3, 1, 2 \rangle \]

### Step 3: Find the direction vector of the line
The direction vector \( \mathbf{b_1} \) of the line \( x = y = z \) can be represented as:
\[ \mathbf{b_1} = \langle 1, 1, 1 \rangle \]

### Step 4: Calculate the cross product of the normal vectors
To find the direction vector \( \mathbf{b_2} \) that is perpendicular to both planes, we calculate the cross product \( \mathbf{b_2} = \mathbf{n_1} \times \mathbf{n_2} \):
\[ \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 1 \\ 3 & 1 & 2 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{b_2} = \mathbf{i} \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} \]
Calculating the minors:
\[ \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = 1, \quad \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} = 4 - 3 = 1, \quad \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = 2 - 3 = -1 \]
Thus:
\[ \mathbf{b_2} = \mathbf{i}(1) - \mathbf{j}(1) + \mathbf{k}(-1) = \langle 1, -1, -1 \rangle \]

### Step 5: Find a point on the line
A point on the line \( x = y = z \) can be taken as \( A(0, 0, 0) \).

### Step 6: Find a point of intersection of the planes
To find the intersection of the planes, we can solve the equations:
1. \( 2x + y + z = 1 \)
2. \( 3x + y + 2z = 2 \)

From the first equation, we can express \( z \):
\[ z = 1 - 2x - y \]
Substituting into the second equation:
\[ 3x + y + 2(1 - 2x - y) = 2 \]
Simplifying:
\[ 3x + y + 2 - 4x - 2y = 2 \implies -x - y = 0 \implies y = -x \]
Substituting \( y = -x \) back into the first equation:
\[ 2x - x + z = 1 \implies z = 1 - x \]
Thus, the point of intersection can be represented as \( B(x, -x, 1 - x) \).

### Step 7: Calculate the shortest distance
The formula for the shortest distance \( d \) between a point and a line is given by:
\[ d = \frac{|\mathbf{b_1} \times \mathbf{b_2} \cdot (\mathbf{A} - \mathbf{B})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]
Calculating \( \mathbf{A} - \mathbf{B} \) where \( \mathbf{B} = (x, -x, 1-x) \):
\[ \mathbf{A} - \mathbf{B} = (0 - x, 0 + x, 0 - (1 - x)) = (-x, x, x - 1) \]

### Step 8: Calculate the cross product \( \mathbf{b_1} \times \mathbf{b_2} \)
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 1 & -1 & -1 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{b_1} \times \mathbf{b_2} = \mathbf{i}(1 \cdot -1 - 1 \cdot -1) - \mathbf{j}(1 \cdot -1 - 1 \cdot 1) + \mathbf{k}(1 \cdot -1 - 1 \cdot 1) \]
\[ = \mathbf{i}(0) - \mathbf{j}(-2) + \mathbf{k}(-2) = \langle 0, 2, -2 \rangle \]

### Step 9: 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 10: Substitute into the distance formula
Now substituting into the distance formula:
\[ d = \frac{|(0, 2, -2) \cdot (-x, x, x - 1)|}{2\sqrt{2}} \]
Calculating the dot product:
\[ = \frac{|0 \cdot -x + 2 \cdot x + (-2)(x - 1)|}{2\sqrt{2}} = \frac{|2x - 2x + 2|}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the shortest distance between the line and the intersection of the planes is:
\[ d = \frac{1}{\sqrt{2}} \]