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: Identify the lines and their vector forms

The first line is given by:
\[ \frac{x-3}{3} = \frac{y-8}{-1} = \frac{z-3}{1} \]
This can be expressed in vector form as:
\[ \mathbf{r_1} = (3, 8, 3) + \lambda (3, -1, 1) \]
where \( \mathbf{a_1} = (3, 8, 3) \) and \( \mathbf{b_1} = (3, -1, 1) \).

The second line is given by:
\[ \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-6}{4} \]
This can be expressed in vector form as:
\[ \mathbf{r_2} = (-3, -7, 6) + \mu (-3, 2, 4) \]
where \( \mathbf{a_2} = (-3, -7, 6) \) and \( \mathbf{b_2} = (-3, 2, 4) \).

### Step 2: Calculate \( \mathbf{a_2} - \mathbf{a_1} \)

Now, we calculate \( \mathbf{a_2} - \mathbf{a_1} \):
\[ \mathbf{a_2} - \mathbf{a_1} = (-3, -7, 6) - (3, 8, 3) = (-6, -15, 3) \]

### Step 3: Calculate \( \mathbf{b_1} \times \mathbf{b_2} \)

Next, we compute the cross product \( \mathbf{b_1} \times \mathbf{b_2} \):
\[ \mathbf{b_1} = (3, -1, 1), \quad \mathbf{b_2} = (-3, 2, 4) \]
Using the determinant to find the cross product:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i}((-1)(4) - (1)(2)) - \mathbf{j}((3)(4) - (1)(-3)) + \mathbf{k}((3)(2) - (-1)(-3)) \]
\[ = \mathbf{i}(-4 - 2) - \mathbf{j}(12 + 3) + \mathbf{k}(6 - 3) \]
\[ = -6\mathbf{i} - 15\mathbf{j} + 3\mathbf{k} \]
Thus,
\[ \mathbf{b_1} \times \mathbf{b_2} = (-6, -15, 3) \]

### Step 4: Calculate the magnitudes

Now, we calculate the magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \):
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} \]

### Step 5: Use the shortest distance formula

The formula for the shortest distance \( d \) between two skew lines is given by:
\[ d = \frac{|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]
Substituting the values we have:
\[ d = \frac{|(-6, -15, 3) \cdot (-6, -15, 3)|}{\sqrt{270}} \]
Calculating the dot product:
\[ (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270 \]
Thus,
\[ d = \frac{270}{\sqrt{270}} = \sqrt{270} \]

### Step 6: Relate to given condition

We know from the problem statement that:
\[ d = \lambda \sqrt{30} \]
Setting these equal gives:
\[ \sqrt{270} = \lambda \sqrt{30} \]

### Step 7: Solve for \( \lambda \)

Squaring both sides:
\[ 270 = \lambda^2 \cdot 30 \]
Thus,
\[ \lambda^2 = \frac{270}{30} = 9 \implies \lambda = 3 \]

### Final Answer

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