If the straight lines `(x-1)/(k)=(y-2)/(2)=(z-3)/(3) and (x-2)/(3)=(y-3)/(k)=(z-1)/(2)` intersect at a point, then the integer `k` is equal to

Options

A. a) `-2`
B. b) `-5`
C. c) `5`
D. d) `2`

Answer

To find the integer value of \( k \) for which the lines

\[ \frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3} \]

and

\[ \frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2} \]

intersect at a point, we can follow these steps:

### Step 1: Parameterize the Lines

Let’s denote the first line by parameter \( t \) and the second line by parameter \( m \).

For the first line:

\[ x = kt + 1, \quad y = 2t + 2, \quad z = 3t + 3 \]

For the second line:

\[ x = 3m + 2, \quad y = km + 3, \quad z = 2m + 1 \]

### Step 2: Set the Coordinates Equal

Since the lines intersect, we can set the coordinates equal to each other:

1. \( kt + 1 = 3m + 2 \) (Equation 1)
2. \( 2t + 2 = km + 3 \) (Equation 2)
3. \( 3t + 3 = 2m + 1 \) (Equation 3)

### Step 3: Solve for \( m \)

From Equation 3, we can express \( m \) in terms of \( t \):

\[ 3t + 3 = 2m + 1 \implies 2m = 3t + 2 \implies m = \frac{3t + 2}{2} \]

### Step 4: Substitute \( m \) into Equations 1 and 2

Substituting \( m \) into Equation 1:

\[ kt + 1 = 3\left(\frac{3t + 2}{2}\right) + 2 \]

This simplifies to:

\[ kt + 1 = \frac{9t + 6}{2} + 2 \implies kt + 1 = \frac{9t + 6 + 4}{2} \implies kt + 1 = \frac{9t + 10}{2} \]

Multiplying through by 2 to eliminate the fraction:

\[ 2kt + 2 = 9t + 10 \implies 2kt - 9t = 8 \implies t(2k - 9) = 8 \implies t = \frac{8}{2k - 9} \tag{Equation 4} \]

### Step 5: Substitute \( m \) into Equation 2

Now substituting \( m \) into Equation 2:

\[ 2t + 2 = k\left(\frac{3t + 2}{2}\right) + 3 \]

This simplifies to:

\[ 2t + 2 = \frac{3kt + 2k}{2} + 3 \]

Multiplying through by 2:

\[ 4t + 4 = 3kt + 2k + 6 \implies 4t - 3kt = 2k + 2 \implies t(4 - 3k) = 2k + 2 \tag{Equation 5} \]

### Step 6: Equate Equations 4 and 5

From Equations 4 and 5, we have:

\[ \frac{8}{2k - 9} = \frac{2k + 2}{4 - 3k} \]

Cross-multiplying gives:

\[ 8(4 - 3k) = (2k + 2)(2k - 9) \]

Expanding both sides:

\[ 32 - 24k = 4k^2 - 18k + 4k - 18 \]

This simplifies to:

\[ 32 - 24k = 4k^2 - 14k - 18 \]

Rearranging gives:

\[ 4k^2 + 10k - 50 = 0 \]

### Step 7: Solve the Quadratic Equation

Dividing the entire equation by 2:

\[ 2k^2 + 5k - 25 = 0 \]

Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 2, b = 5, c = -25 \):

\[ k = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-25)}}{2 \cdot 2} = \frac{-5 \pm \sqrt{25 + 200}}{4} = \frac{-5 \pm 15}{4} \]

Calculating the roots:

1. \( k = \frac{10}{4} = 2.5 \)
2. \( k = \frac{-20}{4} = -5 \)

### Step 8: Determine Integer Value of \( k \)

Since \( k \) must be an integer, we have:

\[ k = -5 \]

### Final Answer

Thus, the integer \( k \) is equal to:

\[ \boxed{-5} \]