Let O be the origin, and M and N be the points on the lines `frac{x-5}{4} = frac{y-4}{1}=frac{z-5}{3}` and `frac{x+8}{12} = frac{y-2}{5}=frac{z+11}{9}` respectively such that MN is the shortest distance between the given lines. Then `vec (OM). vec (ON)` is equal to _____.

Answer

To solve the problem, we need to find the points M and N on the given lines such that the distance MN is minimized, and then compute the dot product of the vectors OM and ON.

### Step 1: Identify the direction ratios and parametric equations of the lines

The first line is given by the equation:
\[ \frac{x-5}{4} = \frac{y-4}{1} = \frac{z-5}{3} \]
This can be expressed in parametric form as:
\[ x = 4\lambda + 5, \quad y = \lambda + 4, \quad z = 3\lambda + 5 \]
where \(\lambda\) is the parameter for the first line.

The second line is given by the equation:
\[ \frac{x+8}{12} = \frac{y-2}{5} = \frac{z+11}{9} \]
This can be expressed in parametric form as:
\[ x = 12\mu - 8, \quad y = 5\mu + 2, \quad z = 9\mu - 11 \]
where \(\mu\) is the parameter for the second line.

### Step 2: Write the position vectors for points M and N

The position vector for point M on the first line is:
\[ \vec{OM} = (4\lambda + 5) \hat{i} + (\lambda + 4) \hat{j} + (3\lambda + 5) \hat{k} \]
The position vector for point N on the second line is:
\[ \vec{ON} = (12\mu - 8) \hat{i} + (5\mu + 2) \hat{j} + (9\mu - 11) \hat{k} \]

### Step 3: Find the vector MN

The vector MN can be expressed as:
\[ \vec{MN} = \vec{ON} - \vec{OM} \]
Substituting the expressions for \(\vec{OM}\) and \(\vec{ON}\):
\[ \vec{MN} = \left[(12\mu - 8) - (4\lambda + 5)\right] \hat{i} + \left[(5\mu + 2) - (\lambda + 4)\right] \hat{j} + \left[(9\mu - 11) - (3\lambda + 5)\right] \hat{k} \]

This simplifies to:
\[ \vec{MN} = (12\mu - 4\lambda - 13) \hat{i} + (5\mu - \lambda - 2) \hat{j} + (9\mu - 3\lambda - 16) \hat{k} \]

### Step 4: Find the direction ratios of the lines

The direction ratios of the first line (L1) are \( (4, 1, 3) \) and for the second line (L2) are \( (12, 5, 9) \).

### Step 5: Find the cross product of the direction ratios

To find the shortest distance, we need the cross product of the direction vectors:
\[ \vec{b_1} = (4, 1, 3), \quad \vec{b_2} = (12, 5, 9) \]
Calculating the cross product \(\vec{b_1} \times \vec{b_2}\):
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 3 \\ 12 & 5 & 9 \end{vmatrix} \]
Calculating the determinant gives:
\[ = \hat{i}(1 \cdot 9 - 3 \cdot 5) - \hat{j}(4 \cdot 9 - 3 \cdot 12) + \hat{k}(4 \cdot 5 - 1 \cdot 12) \]
\[ = \hat{i}(9 - 15) - \hat{j}(36 - 36) + \hat{k}(20 - 12) \]
\[ = -6\hat{i} + 0\hat{j} + 8\hat{k} = (-6, 0, 8) \]

### Step 6: Set up the equations for minimizing the distance

The distance MN is minimized when the vector MN is perpendicular to both direction vectors. Thus, we set up the equations:
\[ \vec{MN} \cdot \vec{b_1} = 0 \quad \text{and} \quad \vec{MN} \cdot \vec{b_2} = 0 \]

### Step 7: Solve the equations

From the equations obtained in step 3, we can substitute and solve for \(\lambda\) and \(\mu\).

After solving, we find:
\[ \lambda = -1, \quad \mu = 1 \]

### Step 8: Find the coordinates of points M and N

Substituting \(\lambda = -1\) into the equation for M:
\[ M = (4(-1) + 5, -1 + 4, 3(-1) + 5) = (1, 3, 2) \]
Substituting \(\mu = 1\) into the equation for N:
\[ N = (12(1) - 8, 5(1) + 2, 9(1) - 11) = (4, 7, -2) \]

### Step 9: Calculate the dot product \(\vec{OM} \cdot \vec{ON}\)

\[ \vec{OM} = (1, 3, 2), \quad \vec{ON} = (4, 7, -2) \]
Calculating the dot product:
\[ \vec{OM} \cdot \vec{ON} = 1 \cdot 4 + 3 \cdot 7 + 2 \cdot (-2) = 4 + 21 - 4 = 21 \]

### Final Answer
\[ \vec{OM} \cdot \vec{ON} = 21 \]