Find the value of `lambda,` so that the lines
`(x-1)/(1)=(y-2)/(2)=(z+lambda)/(3)` and
`(x+1)/(2)=(y-1)/(3)=(z-3)/(1)` are coplanar
Also find the equation of the plane containing them.

Answer

To find the value of \( \lambda \) such that the lines

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

and

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

are coplanar, we will follow these steps:

### Step 1: Identify Points and Direction Ratios

The first line can be expressed in parametric form:
- Point \( A(1, 2, -\lambda) \)
- Direction ratios \( \mathbf{d_1} = (1, 2, 3) \)

The second line can also be expressed in parametric form:
- Point \( B(-1, 1, 3) \)
- Direction ratios \( \mathbf{d_2} = (2, 3, 1) \)

### Step 2: Find Vector \( \mathbf{AB} \)

The vector \( \mathbf{AB} \) from point \( A \) to point \( B \) is given by:
\[ \mathbf{AB} = B - A = (-1 - 1, 1 - 2, 3 - (-\lambda)) = (-2, -1, 3 + \lambda) \]

### Step 3: Set Up the Scalar Triple Product

For the lines to be coplanar, the scalar triple product of the vectors \( \mathbf{AB}, \mathbf{d_1}, \mathbf{d_2} \) must be zero:
\[ \mathbf{AB} \cdot (\mathbf{d_1} \times \mathbf{d_2}) = 0 \]

### Step 4: Calculate \( \mathbf{d_1} \times \mathbf{d_2} \)

To find \( \mathbf{d_1} \times \mathbf{d_2} \), we compute the determinant:

\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ \mathbf{i} \begin{vmatrix} 2 & 3 \\ 3 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} \]

Calculating the minors:

1. \( 2 \cdot 1 - 3 \cdot 3 = 2 - 9 = -7 \)
2. \( 1 \cdot 1 - 3 \cdot 2 = 1 - 6 = -5 \)
3. \( 1 \cdot 3 - 2 \cdot 2 = 3 - 4 = -1 \)

Thus,

\[ \mathbf{d_1} \times \mathbf{d_2} = (-7, 5, -1) \]

### Step 5: Compute the Scalar Triple Product

Now we compute \( \mathbf{AB} \cdot (-7, 5, -1) \):

\[ \mathbf{AB} \cdot (-7, 5, -1) = (-2)(-7) + (-1)(5) + (3 + \lambda)(-1) \]

This simplifies to:

\[ 14 - 5 - (3 + \lambda) = 14 - 5 - 3 - \lambda = 6 - \lambda \]

Setting this equal to zero for coplanarity:

\[ 6 - \lambda = 0 \implies \lambda = 6 \]

### Step 6: Find the Equation of the Plane

To find the equation of the plane containing the lines, we use the normal vector \( \mathbf{n} = (-7, 5, -1) \) and point \( A(1, 2, -6) \):

The equation of the plane is given by:

\[ \mathbf{n} \cdot (\mathbf{r} - \mathbf{A}) = 0 \]

Where \( \mathbf{r} = (x, y, z) \) and \( \mathbf{A} = (1, 2, -6) \):

\[ -7(x - 1) + 5(y - 2) - 1(z + 6) = 0 \]

Expanding this:

\[ -7x + 7 + 5y - 10 - z - 6 = 0 \]

Combining terms:

\[ -7x + 5y - z - 9 = 0 \implies 7x - 5y + z = 9 \]

### Final Results

Thus, the value of \( \lambda \) is \( 6 \) and the equation of the plane containing the lines is:

\[ 7x - 5y + z = 9 \]