If the two lines
`(x-1)/2 = (y+2)/3 = (z-1)/4` and `(x-3)/1 = (y-b)/2 =z/1` intersect each other, then: b=

Options

A. `3/2`
B. `7/2`
C. `-2/9`
D. `-3/2`

Answer

To find the value of \( b \) such that the two lines intersect, we will start by expressing the equations of the lines in parametric form.

### Step 1: Parametric equations of the first line
The first line is given by:
\[ \frac{x-1}{2} = \frac{y+2}{3} = \frac{z-1}{4} \]
Let this common ratio be \( r \). Then we can express \( x, y, z \) in terms of \( r \):
\[ x = 2r + 1 \]
\[ y = 3r - 2 \]
\[ z = 4r + 1 \]

### Step 2: Parametric equations of the second line
The second line is given by:
\[ \frac{x-3}{1} = \frac{y-b}{2} = z \]
Let this common ratio be \( k \). Then we can express \( x, y, z \) in terms of \( k \):
\[ x = k + 3 \]
\[ y = 2k + b \]
\[ z = k \]

### Step 3: Set the equations equal for intersection
For the lines to intersect, the coordinates must be equal at some values of \( r \) and \( k \):
1. \( 2r + 1 = k + 3 \)
2. \( 3r - 2 = 2k + b \)
3. \( 4r + 1 = k \)

### Step 4: Solve the first equation for \( k \)
From the first equation:
\[ k = 2r + 1 - 3 = 2r - 2 \]

### Step 5: Substitute \( k \) into the third equation
Substituting \( k \) into the third equation:
\[ 4r + 1 = 2r - 2 \]
Rearranging gives:
\[ 4r - 2r = -2 - 1 \]
\[ 2r = -3 \implies r = -\frac{3}{2} \]

### Step 6: Find \( k \) using \( r \)
Now substitute \( r \) back to find \( k \):
\[ k = 2\left(-\frac{3}{2}\right) - 2 = -3 - 2 = -5 \]

### Step 7: Substitute \( r \) and \( k \) into the second equation
Now substitute \( r \) and \( k \) into the second equation:
\[ 3\left(-\frac{3}{2}\right) - 2 = 2(-5) + b \]
Calculating the left side:
\[ -\frac{9}{2} - 2 = -\frac{9}{2} - \frac{4}{2} = -\frac{13}{2} \]
So we have:
\[ -\frac{13}{2} = -10 + b \]
Rearranging gives:
\[ b = -\frac{13}{2} + 10 = -\frac{13}{2} + \frac{20}{2} = \frac{7}{2} \]

### Final Answer
Thus, the value of \( b \) is:
\[ \boxed{\frac{7}{2}} \]