A dimensionless quantity is constructed in terms of electronic charge $e$, permittivity of free space $\varepsilon_0$, Planck's constant $h$, and speed of light $c$. If the dimensionless quantity is written as $e^{\alpha}\varepsilon_0^{\beta}h^{\gamma}c^{\delta}$ and $n$ is a non-zero integer, then $(\alpha, \beta, \gamma, \delta)$ is given by

Step 1: Write down the dimensions of each physical quantity.

The dimensions of electronic charge $e$, speed of light $c$, Planck's constant $h$, and permittivity of free space $\varepsilon_0$ are given by:

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

Step 2: Set up the dimensionless condition.

The quantity $e^{\alpha}\varepsilon_0^{\beta}h^{\gamma}c^{\delta}$ must be dimensionless.

$$[e^{\alpha}\varepsilon_0^{\beta}h^{\gamma}c^{\delta}] = \text{M}^{0}\,\text{L}^{0}\,\text{T}^{0}\,\text{A}^{0}$$

Substitute the dimensions of each quantity into the expression.

$$(\text{A}\,\text{T})^{\alpha} (\text{M}^{-1}\,\text{L}^{-3}\,\text{T}^{4}\,\text{A}^{2})^{\beta} (\text{M}\,\text{L}^{2}\,\text{T}^{-1})^{\gamma} (\text{L}\,\text{T}^{-1})^{\delta} = \text{M}^{0}\,\text{L}^{0}\,\text{T}^{0}\,\text{A}^{0}$$

Step 3: Equate the exponents of each base dimension.

Equating the powers of $\text{M}$, $\text{L}$, $\text{T}$, and $\text{A}$ on both sides yields a system of linear equations.

For $\text{A}$:

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

For $\text{M}$:

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

For $\text{L}$:

$$-3\beta + 2\gamma + \delta = 0$$

For $\text{T}$:

$$\alpha + 4\beta - \gamma - \delta = 0$$

Step 4: Solve the system of equations.

Express $\alpha$, $\gamma$, and $\delta$ in terms of $\beta$.

From the equation for $\text{A}$:

$$\alpha = -2\beta$$

From the equation for $\text{M}$:

$$\gamma = \beta$$

Substitute $\gamma = \beta$ into the equation for $\text{L}$:

$$-3\beta + 2\beta + \delta = 0$$ $$\delta = \beta$$

Verify these relations in the equation for $\text{T}$.

$$-2\beta + 4\beta - \beta - \beta = 0$$ $$0 = 0$$

Step 5: Express the solution in terms of the non-zero integer $n$.

Let $\beta = -n$, where $n$ is a non-zero integer.

$$\alpha = 2n$$ $$\beta = -n$$ $$\gamma = -n$$ $$\delta = -n$$

This gives the ordered quadruple.

$$(\alpha, \beta, \gamma, \delta) = (2n, -n, -n, -n)$$

Answer: $(\alpha, \beta, \gamma, \delta) = (2n, -n, -n, -n)$