If the lines `(x)/(1)=(y)/(2)=(z)/(3), (x-k)/(3)=(y-3)/(-1)=(z-4)/(h)` and `(2x+1)/(3)=(y-1)/(1)=(z-2)/(1)` are concurrent, then the value of `2h-3k` is equal to

Options

A. 3
B. 2
C. `-4`
D. 4

Answer

To solve the problem, we need to determine the value of \(2h - 3k\) given that the lines are concurrent. Let's break down the solution step by step.

### Step 1: Parametrize the lines

1. The first line is given by:
\[ \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \]
Let \( \lambda \) be the parameter. Then we can express the coordinates as:
\[ x = \lambda, \quad y = 2\lambda, \quad z = 3\lambda \]

2. The second line is given by:
\[ \frac{x - k}{3} = \frac{y - 3}{-1} = \frac{z - 4}{h} \]
Let \( \mu \) be the parameter. Then we can express the coordinates as:
\[ x = 3\mu + k, \quad y = -\mu + 3, \quad z = h\mu + 4 \]

3. The third line is given by:
\[ \frac{2x + 1}{3} = \frac{y - 1}{1} = \frac{z - 2}{1} \]
Let \( \nu \) be the parameter. Then we can express the coordinates as:
\[ x = \frac{3\nu - 1}{2}, \quad y = \nu + 1, \quad z = \nu + 2 \]

### Step 2: Set equations for concurrency

Since the lines are concurrent, the points from the first two lines must satisfy the equations of the third line.

### Step 3: Equate the coordinates

1. From the second line:
\[ 3\mu + k = \frac{3\nu - 1}{2} \quad (1) \]
\[ -\mu + 3 = \nu + 1 \quad (2) \]
\[ h\mu + 4 = \nu + 2 \quad (3) \]

### Step 4: Solve the equations

From equation (2):
\[ -\mu + 3 = \nu + 1 \implies \nu = 2 - \mu \quad (4) \]

Substituting (4) into (1):
\[ 3\mu + k = \frac{3(2 - \mu) - 1}{2} \]
\[ 3\mu + k = \frac{6 - 3\mu - 1}{2} \]
\[ 3\mu + k = \frac{5 - 3\mu}{2} \]
Multiplying through by 2:
\[ 6\mu + 2k = 5 - 3\mu \]
\[ 9\mu + 2k = 5 \quad (5) \]

Now substituting (4) into (3):
\[ h\mu + 4 = (2 - \mu) + 2 \]
\[ h\mu + 4 = 4 - \mu \]
\[ h\mu + \mu = 0 \implies (h + 1)\mu = 0 \]
This implies either \( \mu = 0 \) or \( h = -1 \).

### Step 5: Find \(k\) and \(h\)

1. If \( \mu = 0 \):
From (4):
\[ \nu = 2 \]
Substitute \( \mu = 0 \) in (5):
\[ 2k = 5 \implies k = \frac{5}{2} \]

2. If \( h = -1 \):
Substitute \( h = -1 \) back into (5):
\[ 9\mu + 2(-1) = 5 \implies 9\mu - 2 = 5 \implies 9\mu = 7 \implies \mu = \frac{7}{9} \]
Substitute \( \mu = \frac{7}{9} \) into (4):
\[ \nu = 2 - \frac{7}{9} = \frac{11}{9} \]
Substitute \( \mu = \frac{7}{9} \) into (5):
\[ 9 \cdot \frac{7}{9} + 2k = 5 \implies 7 + 2k = 5 \implies 2k = -2 \implies k = -1 \]

### Step 6: Calculate \(2h - 3k\)

Using \( h = -1 \) and \( k = -1 \):
\[ 2h - 3k = 2(-1) - 3(-1) = -2 + 3 = 1 \]

### Final Answer
The value of \(2h - 3k\) is \(1\).