<s>[INST] (High School Vectors) Distance between Parallel Lines Given $2$ lines
$$L_1 : r = (1, 2, 3) + \lambda (1, 2, 1.5)\\
L_2 : r = (4, 0, -1) + \mu (2, 4, 3)$$
Are parallel to each other with points $A(1, 2, 3)$ and $B(4, 0, -1)$.
Determine the distance between the two lines.

Method 1: distance from a point B to line L1
$AB(\to) = OB(\to) - OA(\to) = (3, -2, -4)$
the distance is
$$\frac{|(3, -2, -4) \times (1, 2, 1.5) |}{\sqrt{1 + 4 + 2.25}}= 4.716$$

Method 2: foot of perpendicular
Let foot of perpendicular of $A$ on $L2$ be $F$
$$OF(→) = (4+2\mu, 4\mu, -1+3\mu) \text{ for some }\mu\in\mathbb R\\
AF(→) = (3+2\mu, 4\mu-2, -4+3\mu) $$
Since $AF$ perpendicular to $L2$ :
$$(3+2\mu, 4\mu-2, -4+3\mu) \cdot (2, 4, 3) = 0\\
μ = \frac{14}{29}\\
AF(\to) = \left(\frac{115}{29}, \frac{56}{29}, -\frac{74}{29}\right)\\
|AF(\to)| = 5.095$$
Method 3: Length of projection and Pythagoras Theorem 
Length of projection of vector $\vec{AB}$ onto $L_1$:
$$\frac{|(3, -2, -4)\cdot (1, 2, 1.5)|}{\sqrt{1 + 4 + 2.25}} = 2.90065391$$
Length of $\vec{AB} = | (3, -2, -4) |$.
Distance between two lines by Pythagoras theorem: 
$$\sqrt{| (3, -2, -4) |^2 - 2.90065391^2 } = 4.537237 . $$
Why do I get different answers for methods? I believe the three methods are correct. 
 [/INST] The vector $AF$ should be $$\left(\frac{115}{29}, -\frac{2}{29}, -\frac{74}{29}\right)$$
This vector has a norm of approximately $4.716$ which is what you hot with method 1.
Your steps were correct up to the point where you inserted $\mu$ into the formula
$$AF = (3+2\mu, 4\mu - 2, -4+3\mu)$$
since
$$3+2\cdot \frac{14}{29} = \frac{87 + 28}{29} = \frac{115}{29}\\
4\cdot \frac{14}{29} - 2 = \frac{56 - 58}{29} = -\frac{2}{29}\\
-4+3\cdot\frac{14}{29} = \frac{-116 + 42}{29} = -\frac{74}{29}$$
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