If the shortest distance between the lines `(x-3)/(3)=(y-8)/(-1)=(z-3)/(1) and (x+3)/(-3)=(y+7)/(2)=(z-6)/(4) is lambdasqrt(30)` unit, then the value of `lambda` is

Answer

To find the value of \( \lambda \) such that the shortest distance between the given lines is \( \lambda \sqrt{30} \), we will follow these steps:

### Step 1: Write the equations of the lines in vector form
The first line can be expressed as:
\[ \frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1} \]
This can be written in vector form as:
\[ \mathbf{r_1} = \begin{pmatrix} 3 \\ 8 \\ 3 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} \]
where \( t \) is a parameter.

The second line can be expressed as:
\[ \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4} \]
This can be written in vector form as:
\[ \mathbf{r_2} = \begin{pmatrix} -3 \\ -7 \\ 6 \end{pmatrix} + s \begin{pmatrix} -3 \\ 2 \\ 4 \end{pmatrix} \]
where \( s \) is another parameter.

### Step 2: Identify the direction vectors and points
From the equations, we identify:
- Point on the first line \( \mathbf{a_1} = \begin{pmatrix} 3 \\ 8 \\ 3 \end{pmatrix} \) and direction vector \( \mathbf{b_1} = \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} \).
- Point on the second line \( \mathbf{a_2} = \begin{pmatrix} -3 \\ -7 \\ 6 \end{pmatrix} \) and direction vector \( \mathbf{b_2} = \begin{pmatrix} -3 \\ 2 \\ 4 \end{pmatrix} \).

### Step 3: Calculate the vector between the two points
The vector connecting the two points is:
\[ \mathbf{a_1} - \mathbf{a_2} = \begin{pmatrix} 3 \\ 8 \\ 3 \end{pmatrix} - \begin{pmatrix} -3 \\ -7 \\ 6 \end{pmatrix} = \begin{pmatrix} 6 \\ 15 \\ -3 \end{pmatrix} \]

### Step 4: Calculate the cross product of the direction vectors
Now, we need to compute \( \mathbf{b_1} \times \mathbf{b_2} \):
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} \times \begin{pmatrix} -3 \\ 2 \\ 4 \end{pmatrix} \]
Using the determinant method:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = \mathbf{i}((-1)(4) - (1)(2)) - \mathbf{j}((3)(4) - (1)(-3)) + \mathbf{k}((3)(2) - (-1)(-3)) \]
Calculating this gives:
\[ = \mathbf{i}(-4 - 2) - \mathbf{j}(12 + 3) + \mathbf{k}(6 - 3) = -6\mathbf{i} - 15\mathbf{j} + 3\mathbf{k} = \begin{pmatrix} -6 \\ -15 \\ 3 \end{pmatrix} \]

### Step 5: Calculate the magnitude of the cross product
Now, we find the magnitude:
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30} \]

### Step 6: Use the distance formula
The shortest distance \( D \) between the two lines is given by:
\[ D = \frac{|\mathbf{a_1} - \mathbf{a_2} \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]
Calculating the dot product:
\[ |\mathbf{a_1} - \mathbf{a_2}| \cdot (\mathbf{b_1} \times \mathbf{b_2}) = \begin{pmatrix} 6 \\ 15 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} -6 \\ -15 \\ 3 \end{pmatrix} = 6(-6) + 15(-15) + (-3)(3) = -36 - 225 - 9 = -270 \]
Thus, the distance becomes:
\[ D = \frac{|-270|}{3\sqrt{30}} = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{30} \]

### Step 7: Relate to given distance
We know from the problem that this distance is equal to \( \lambda \sqrt{30} \). Therefore, we have:
\[ 3\sqrt{30} = \lambda \sqrt{30} \]
Dividing both sides by \( \sqrt{30} \) gives:
\[ \lambda = 3 \]

### Final Answer
The value of \( \lambda \) is \( \boxed{3} \).