To find the shortest distance between the lines given by the equations \( x + a = 2y = -12z \) and \( x = y + 2a = 6z - 6a \), we will follow these steps:
### Step 1: Rewrite the equations in symmetric form
We start with the equations of the lines:
1. For the first line:
\[
x + a = 2y = -12z
\]
We can express this in symmetric form by dividing each part by \(-12\):
\[
\frac{x + a}{-12} = \frac{y}{-6} = \frac{z}{1}
\]
This gives us the symmetric form of the first line:
\[
\frac{x + a}{-12} = \frac{y}{-6} = z
\]
2. For the second line:
\[
x = y + 2a = 6z - 6a
\]
We can express this in symmetric form by dividing each part by \(6\):
\[
\frac{x}{6} = \frac{y + 2a}{6} = \frac{z - a}{1}
\]
This gives us the symmetric form of the second line:
\[
\frac{x}{6} = \frac{y + 2a}{6} = z - a
\]
### Step 2: Identify direction ratios and points
From the symmetric forms, we can identify the direction ratios and points on each line:
- For the first line, the direction ratios are \((-12, -6, 1)\) and a point on the line can be taken as \((-a, 0, 0)\).
- For the second line, the direction ratios are \((6, 6, 1)\) and a point on the line can be taken as \((0, -2a, a)\).
### Step 3: Use the formula for shortest distance between two skew lines
The formula for the shortest distance \(D\) between two skew lines given by points \(P_1\) and \(P_2\) and direction ratios \( \vec{d_1} \) and \( \vec{d_2} \) is:
\[
D = \frac{|(\vec{P_2} - \vec{P_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}
\]
### Step 4: Calculate the cross product of direction ratios
Let \( \vec{d_1} = (-12, -6, 1) \) and \( \vec{d_2} = (6, 6, 1) \).
Calculating the cross product \( \vec{d_1} \times \vec{d_2} \):
\[
\vec{d_1} \times \vec{d_2} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-12 & -6 & 1 \\
6 & 6 & 1
\end{vmatrix}
\]
Calculating the determinant:
\[
= \hat{i}((-6)(1) - (1)(6)) - \hat{j}((-12)(1) - (1)(6)) + \hat{k}((-12)(6) - (-6)(6))
\]
\[
= \hat{i}(-6 - 6) - \hat{j}(-12 - 6) + \hat{k}(-72 + 36)
\]
\[
= \hat{i}(-12) + \hat{j}(18) + \hat{k}(-36)
\]
Thus, \( \vec{d_1} \times \vec{d_2} = (-12, 18, -36) \).
### Step 5: Calculate the magnitude of the cross product
\[
|\vec{d_1} \times \vec{d_2}| = \sqrt{(-12)^2 + 18^2 + (-36)^2} = \sqrt{144 + 324 + 1296} = \sqrt{1764} = 42
\]
### Step 6: Calculate \( \vec{P_2} - \vec{P_1} \)
Let \( \vec{P_1} = (-a, 0, 0) \) and \( \vec{P_2} = (0, -2a, a) \):
\[
\vec{P_2} - \vec{P_1} = (0 + a, -2a - 0, a - 0) = (a, -2a, a)
\]
### Step 7: Calculate the dot product
Now, we calculate the dot product:
\[
(\vec{P_2} - \vec{P_1}) \cdot (\vec{d_1} \times \vec{d_2}) = (a, -2a, a) \cdot (-12, 18, -36)
\]
\[
= a(-12) + (-2a)(18) + a(-36) = -12a - 36a - 36a = -84a
\]
### Step 8: Calculate the shortest distance
Now, substituting back into the distance formula:
\[
D = \frac{| -84a |}{42} = \frac{84a}{42} = 2a
\]
Thus, the shortest distance between the two lines is \( \boxed{2a} \).
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