If the line `r=(i+j-k)+lambda(3i-j)andr=(4i-k)+mu(2i+3k)` intersect at the point `(p,0,p-5)` then

Options

A. `p=0`
B. `p=-1`
C. `p=4`
D. `p=5`

Answer

To solve the problem, we need to find the value of \( p \) where the two lines intersect at the point \( (p, 0, p-5) \).

### Step 1: Write the equations of the lines in vector form

The first line is given by:
\[ \mathbf{r_1} = (1, 1, -1) + \lambda(3, -1, 0) \]
This can be expressed in component form as:
\[ x_1 = 1 + 3\lambda, \quad y_1 = 1 - \lambda, \quad z_1 = -1 \]

The second line is given by:
\[ \mathbf{r_2} = (4, 0, -1) + \mu(2, 0, 3) \]
This can be expressed in component form as:
\[ x_2 = 4 + 2\mu, \quad y_2 = 0, \quad z_2 = -1 + 3\mu \]

### Step 2: Set the equations equal to find the intersection

Since the lines intersect at the point \( (p, 0, p-5) \), we can set the corresponding components equal to each other.

1. For the \( x \)-coordinates:
\[ 1 + 3\lambda = p \tag{1} \]
2. For the \( y \)-coordinates:
\[ 1 - \lambda = 0 \tag{2} \]
3. For the \( z \)-coordinates:
\[ -1 = -1 + 3\mu \tag{3} \]

### Step 3: Solve the equations

**From equation (2)**:
\[ 1 - \lambda = 0 \implies \lambda = 1 \]

**Substituting \(\lambda = 1\) into equation (1)**:
\[ 1 + 3(1) = p \implies p = 4 \]

**From equation (3)**:
\[ -1 = -1 + 3\mu \implies 0 = 3\mu \implies \mu = 0 \]

### Step 4: Verify the intersection point

Now we can verify that both lines intersect at the point \( (4, 0, 4-5) = (4, 0, -1) \).

**For the first line** with \(\lambda = 1\):
\[ x_1 = 1 + 3(1) = 4, \quad y_1 = 1 - 1 = 0, \quad z_1 = -1 \]
This gives the point \( (4, 0, -1) \).

**For the second line** with \(\mu = 0\):
\[ x_2 = 4 + 2(0) = 4, \quad y_2 = 0, \quad z_2 = -1 + 3(0) = -1 \]
This also gives the point \( (4, 0, -1) \).

### Conclusion

Both lines intersect at the point \( (4, 0, -1) \), confirming that \( p = 4 \).

Thus, the value of \( p \) is:
\[ \boxed{4} \]