The equation of the plane passing through the poit of intersection of the lines `(x-1)/(3)=(y-2)/(1)=(z-3)/(2),(x-3)/(1)=(y-1)/(2)=(z-2)/(3)` and perpendicular to the line `(x-2)/(2)=(y-3)/(3)=(z-2)/(1)` is P = 0. If the distance of the point `(1, 1, 3)` from P = 0 is k units, then the value of `(k^(2))/(2)` is equal to

Answer

To solve the problem step by step, we need to find the equation of the plane that passes through the intersection of two given lines and is perpendicular to a third line. We will then calculate the distance from a given point to this plane and find the value of \( \frac{k^2}{2} \).

### Step 1: Find the parametric equations of the lines

The first line \( L_1 \) is given by:
\[ \frac{x-1}{3} = \frac{y-2}{1} = \frac{z-3}{2} = \alpha \]
From this, we can express the coordinates in terms of \( \alpha \):
\[ x = 3\alpha + 1, \quad y = \alpha + 2, \quad z = 2\alpha + 3 \]

The second line \( L_2 \) is given by:
\[ \frac{x-3}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \beta \]
From this, we can express the coordinates in terms of \( \beta \):
\[ x = \beta + 3, \quad y = 2\beta + 1, \quad z = 3\beta + 2 \]

### Step 2: Find the point of intersection of the lines

To find the intersection, we set the \( x \) coordinates equal:
\[ 3\alpha + 1 = \beta + 3 \]
This simplifies to:
\[ 3\alpha - \beta = 2 \quad \text{(1)} \]

Next, we set the \( y \) coordinates equal:
\[ \alpha + 2 = 2\beta + 1 \]
This simplifies to:
\[ \alpha - 2\beta = -1 \quad \text{(2)} \]

Now we have a system of equations:
1. \( 3\alpha - \beta = 2 \)
2. \( \alpha - 2\beta = -1 \)

### Step 3: Solve the system of equations

From equation (2), we can express \( \alpha \) in terms of \( \beta \):
\[ \alpha = 2\beta - 1 \]

Substituting this into equation (1):
\[ 3(2\beta - 1) - \beta = 2 \]
This simplifies to:
\[ 6\beta - 3 - \beta = 2 \implies 5\beta = 5 \implies \beta = 1 \]

Now substitute \( \beta = 1 \) back into the expression for \( \alpha \):
\[ \alpha = 2(1) - 1 = 1 \]

### Step 4: Find the coordinates of the intersection point

Substituting \( \alpha = 1 \) into the equations for \( L_1 \):
\[ x = 3(1) + 1 = 4, \quad y = 1 + 2 = 3, \quad z = 2(1) + 3 = 5 \]
Thus, the point of intersection is \( (4, 3, 5) \).

### Step 5: Find the normal vector of the plane

The direction ratios of the line that the plane is perpendicular to are given by:
\[ \frac{x-2}{2} = \frac{y-3}{3} = \frac{z-2}{1} \]
Thus, the direction ratios are \( (2, 3, 1) \). Therefore, the normal vector \( \vec{n} \) of the plane is:
\[ \vec{n} = 2\hat{i} + 3\hat{j} + 1\hat{k} \]

### Step 6: Write the equation of the plane

Using the point-normal form of the equation of a plane:
\[ 2(x - 4) + 3(y - 3) + 1(z - 5) = 0 \]
Expanding this gives:
\[ 2x - 8 + 3y - 9 + z - 5 = 0 \implies 2x + 3y + z - 22 = 0 \]

### Step 7: Calculate the distance from the point \( (1, 1, 3) \) to the plane

The distance \( D \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For our plane \( 2x + 3y + z - 22 = 0 \), we have \( A = 2, B = 3, C = 1, D = -22 \) and the point \( (1, 1, 3) \):
\[ D = \frac{|2(1) + 3(1) + 1(3) - 22|}{\sqrt{2^2 + 3^2 + 1^2}} = \frac{|2 + 3 + 3 - 22|}{\sqrt{14}} = \frac{|-14|}{\sqrt{14}} = \frac{14}{\sqrt{14}} = \sqrt{14} \]

### Step 8: Find \( k^2/2 \)

Since \( k = \sqrt{14} \), we have:
\[ k^2 = 14 \implies \frac{k^2}{2} = \frac{14}{2} = 7 \]

Thus, the final answer is:
\[ \boxed{7} \]