For two vectors $\vec{A}$ and $\vec{B}$ to be parallel, their cross product must be zero, or equivalently, their corresponding components must be proportional.

The given vectors are:

$$\begin{aligned} \vec{A} &= 5\hat{i} + 2\hat{j} - 3\hat{k} \\ \vec{B} &= 15\hat{i} + a\hat{j} - 9\hat{k} \end{aligned}$$

Method 1: Using Component Proportionality

Comparing the components of $\vec{A}$ and $\vec{B}$:

$$\begin{aligned} A_x &= 5, \quad A_y = 2, \quad A_z = -3 \\ B_x &= 15, \quad B_y = a, \quad B_z = -9 \end{aligned}$$

For the vectors to be parallel, the ratio of their corresponding components must be equal:

$$\frac{A_x}{B_x} = \frac{A_y}{B_y} = \frac{A_z}{B_z}$$

Substituting the values:

$$\frac{5}{15} = \frac{2}{a} = \frac{-3}{-9}$$

Simplifying the known ratios:

$$\frac{1}{3} = \frac{2}{a} = \frac{1}{3}$$

Now, solving for $a$:

$$\begin{aligned} \frac{2}{a} &= \frac{1}{3} \\ a &= 2 \times 3 \\ a &= 6 \end{aligned}$$

Method 2: Using the Vector Cross Product

For parallel vectors, the cross product is zero:

$$\vec{A} \times \vec{B} = \vec{0}$$

We can express the cross product as a determinant:

$$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 2 & -3 \\ 15 & a & -9 \end{vmatrix} = \vec{0}$$

Expanding the determinant:

$$\begin{aligned} \vec{A} \times \vec{B} &= \hat{i} \begin{vmatrix} 2 & -3 \\ a & -9 \end{vmatrix} - \hat{j} \begin{vmatrix} 5 & -3 \\ 15 & -9 \end{vmatrix} + \hat{k} \begin{vmatrix} 5 & 2 \\ 15 & a \end{vmatrix} \\ &= \hat{i} [2(-9) - (-3)(a)] - \hat{j} [5(-9) - (-3)(15)] + \hat{k} [5(a) - 2(15)] \\ &= \hat{i} (-18 + 3a) - \hat{j} (-45 + 45) + \hat{k} (5a - 30) \\ &= (3a - 18)\hat{i} + (5a - 30)\hat{k} \end{aligned}$$

Since $\vec{A} \times \vec{B} = 0\hat{i} + 0\hat{j} + 0\hat{k}$, we equate the components to zero:

$$\begin{aligned} 3a - 18 &= 0 \implies 3a = 18 \implies a = 6 \\ 5a - 30 &= 0 \implies 5a = 30 \implies a = 6 \end{aligned}$$

Both methods yield $a = 6$. Thus, the correct option is A.