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 will follow the approach outlined in the video transcript.

### Step 1: Find the Point of Intersection of the Two Lines

The equations of the lines are given as:
1. \(\frac{x-1}{3} = \frac{y-2}{1} = \frac{z-3}{2} = \lambda\)
2. \(\frac{x-3}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \mu\)

Let's express the coordinates of points on both lines in terms of \(\lambda\) and \(\mu\):

For Line 1 (L1):
- \(x = 3\lambda + 1\)
- \(y = \lambda + 2\)
- \(z = 2\lambda + 3\)

For Line 2 (L2):
- \(x = \mu + 3\)
- \(y = 2\mu + 1\)
- \(z = 3\mu + 2\)

Setting the coordinates equal to find the intersection:
1. \(3\lambda + 1 = \mu + 3\)
2. \(\lambda + 2 = 2\mu + 1\)
3. \(2\lambda + 3 = 3\mu + 2\)

### Step 2: Solve the System of Equations

From the first equation:
\[ 3\lambda - \mu = 2 \quad \text{(1)} \]

From the second equation:
\[ \lambda - 2\mu = -1 \quad \text{(2)} \]

Now, we can multiply equation (2) by 3:
\[ 3\lambda - 6\mu = -3 \quad \text{(3)} \]

Now we have:
1. \(3\lambda - \mu = 2\) (1)
2. \(3\lambda - 6\mu = -3\) (3)

Subtract equation (1) from equation (3):
\[ (3\lambda - 6\mu) - (3\lambda - \mu) = -3 - 2 \]
\[ -5\mu = -5 \implies \mu = 1 \]

Substituting \(\mu = 1\) into equation (1):
\[ 3\lambda - 1 = 2 \implies 3\lambda = 3 \implies \lambda = 1 \]

### Step 3: Find the Coordinates of the Point of Intersection

Substituting \(\lambda = 1\) into the equations for Line 1:
- \(x = 3(1) + 1 = 4\)
- \(y = 1 + 2 = 3\)
- \(z = 2(1) + 3 = 5\)

Thus, the point of intersection is \(P(4, 3, 5)\).

### Step 4: Find the Normal Vector of the Plane

The plane is perpendicular to the line given by:
\(\frac{x-2}{2} = \frac{y-3}{3} = \frac{z-2}{1}\)

The direction ratios (normal vector) of this line are \(2, 3, 1\).

### Step 5: Write the Equation of the Plane

The equation of the plane can be expressed as:
\[ 2(x - 4) + 3(y - 3) + 1(z - 5) = 0 \]

Expanding this:
\[ 2x - 8 + 3y - 9 + z - 5 = 0 \]
\[ 2x + 3y + z - 22 = 0 \]

### Step 6: Find 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:
\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

Here, \(A = 2\), \(B = 3\), \(C = 1\), and \(D = -22\). The point is \((1, 1, 3)\).

Calculating:
\[ D = \frac{|2(1) + 3(1) + 1(3) - 22|}{\sqrt{2^2 + 3^2 + 1^2}} \]
\[ = \frac{|2 + 3 + 3 - 22|}{\sqrt{4 + 9 + 1}} = \frac{|8 - 22|}{\sqrt{14}} = \frac{14}{\sqrt{14}} = \sqrt{14} \]

### Step 7: Calculate \(k^2 / 2\)

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

### Final Answer
The value of \(\frac{k^2}{2}\) is \(7\).