To find the values of $(\alpha, \beta, \gamma, \delta)$ for which the expression $e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta$ is dimensionless, we need to ensure that the product of the quantities has no overall physical dimensions. This means the dimensions of charge ($e$), permittivity ($\varepsilon_0$), Planck's constant ($h$), and speed of light ($c$) need to cancel each other out.

Let's look at the dimensions of each of the physical quantities:

Electronic charge ($e$): The dimension of charge in terms of fundamental dimensions is:

$$[e] = [A \cdot T]$$

where $A$ is the dimension of current and $T$ is the dimension of time.

Permittivity of free space ($\varepsilon_0$): The dimension is given by:

$$[\varepsilon_0] = [\text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2]$$

Planck's constant ($h$): The dimension is:

$$[h] = [\text{M} \text{L}^2 \text{T}^{-1}]$$

Speed of light ($c$): The dimension is:

$$[c] = [\text{L} \text{T}^{-1}]$$

Now, we write the expression for the dimensionless quantity and set its dimensions to zero:

$$[e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta] = [A^\alpha T^\alpha] [\text{M}^{-\beta} \text{L}^{-3\beta} \text{T}^{4\beta} \text{A}^{2\beta}] [\text{M}^\gamma \text{L}^{2\gamma} \text{T}^{-\gamma}] [\text{L}^\delta \text{T}^{-\delta}]$$

Combining the dimensions, we get:

$$[e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta] = [\text{M}^{-\beta + \gamma} \text{L}^{-3\beta + 2\gamma + \delta} \text{T}^{\alpha + 4\beta - \gamma - \delta} \text{A}^{\alpha + 2\beta}]$$

For the quantity to be dimensionless, the exponents of $\text{M}$, $\text{L}$, $\text{T}$, and $\text{A}$ must all be zero:

• $$-\beta + \gamma = 0$$
• $$-3\beta + 2\gamma + \delta = 0$$
• $$\alpha + 4\beta - \gamma - \delta = 0$$
• $$\alpha + 2\beta = 0$$

Solving the above set of equations, we start with the fourth equation:

$$\alpha + 2\beta = 0 \Rightarrow \alpha = -2\beta$$

Substituting $\alpha = -2\beta$ into the third equation, we get:

$$-2\beta + 4\beta - \gamma - \delta = 0 \Rightarrow 2\beta - \gamma - \delta = 0 \Rightarrow \gamma + \delta = 2\beta$$

From the first equation:

$$-\beta + \gamma = 0 \Rightarrow \gamma = \beta$$

Then, substituting $\gamma = \beta$ into $\gamma + \delta = 2\beta$:

$$\beta + \delta = 2\beta \Rightarrow \delta = \beta$$

Thus, we have:

$$\alpha = -2\beta$$

$$\gamma = \beta$$

$$\delta = \beta$$

Let $\beta = -n$ where $n$ is a non-zero integer. Then, we can write the values as:

$$\alpha = -2(-n) = 2n$$

$$\beta = -n$$

$$\gamma = -n$$

$$\delta = -n$$

Hence, the tuple $(\alpha, \beta, \gamma, \delta)$ is given by:

$$ (2n, -n, -n, -n) $$

So, the correct option is Option A: $(2n,-n,-n,-n)$