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. `-2`
B. `-5`
C. `5`
D. `2`

Answer

To find the integer value of \( k \) such that the given lines intersect at a point, we can follow these steps:

### Step 1: Write the equations of the lines
The equations of the lines are given as:
1. Line \( l_1: \frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3} \)
2. Line \( l_2: \frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2} \)

### Step 2: Identify the direction ratios
From the equations, we can identify the direction ratios of the lines:
- For line \( l_1 \), the direction ratios are \( (k, 2, 3) \).
- For line \( l_2 \), the direction ratios are \( (3, k, 2) \).

### Step 3: Use the condition for coplanarity
For the lines to intersect, they must be coplanar. The condition for coplanarity of two lines with direction ratios \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \) and passing through points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is given by the determinant:

\[ \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 \]

### Step 4: Substitute the points and direction ratios
Let’s take points from the lines:
- For line \( l_1 \): Point \( (1, 2, 3) \)
- For line \( l_2 \): Point \( (2, 3, 1) \)

Now we can calculate:
- \( x_2 - x_1 = 2 - 1 = 1 \)
- \( y_2 - y_1 = 3 - 2 = 1 \)
- \( z_2 - z_1 = 1 - 3 = -2 \)

Thus, the determinant becomes:

\[ \begin{vmatrix} 1 & 1 & -2 \\ k & 2 & 3 \\ 3 & k & 2 \end{vmatrix} = 0 \]

### Step 5: Calculate the determinant
Calculating the determinant:

\[ 1 \cdot \begin{vmatrix} 2 & 3 \\ k & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} k & 3 \\ 3 & 2 \end{vmatrix} - 2 \cdot \begin{vmatrix} k & 2 \\ 3 & k \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 3 \\ k & 2 \end{vmatrix} = 2 \cdot 2 - 3 \cdot k = 4 - 3k \)

2. \( \begin{vmatrix} k & 3 \\ 3 & 2 \end{vmatrix} = k \cdot 2 - 3 \cdot 3 = 2k - 9 \)

3. \( \begin{vmatrix} k & 2 \\ 3 & k \end{vmatrix} = k^2 - 6 \)

Putting it all together:

\[ 1(4 - 3k) - 1(2k - 9) - 2(k^2 - 6) = 0 \]

### Step 6: Simplify the equation
Expanding and simplifying:

\[ 4 - 3k - 2k + 9 - 2k^2 + 12 = 0 \]

Combining like terms:

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

Multiplying through by -1:

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

### Step 7: Solve the quadratic equation
Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

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

\[ k = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-25)}}{2 \cdot 2} \]
\[ k = \frac{-5 \pm \sqrt{25 + 200}}{4} \]
\[ k = \frac{-5 \pm \sqrt{225}}{4} \]
\[ k = \frac{-5 \pm 15}{4} \]

Calculating the two possible values:

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

### Step 8: Identify the integer value of \( k \)
The integer value of \( k \) from the options is:

\[ k = -5 \]

Thus, the integer \( k \) is equal to \( -5 \).

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