Step 1: Find the point of intersection of lines $4x + 7y - 3 = 0$ and $2x - 3y + 1 = 0$.

Solving by cross-multiplication:

$$\frac{x}{7(1) - (-3)(-3)} = \frac{y}{(-3)(2) - 4(1)} = \frac{1}{4(-3) - 7(2)}$$ $$\frac{x}{7-9} = \frac{y}{-6-4} = \frac{1}{-12-14}$$ $$\frac{x}{-2} = \frac{y}{-10} = \frac{1}{-26}$$

Therefore, $x = \frac{1}{13}$ and $y = \frac{5}{13}$.

Point of intersection is $\left(\frac{1}{13}, \frac{5}{13}\right)$.

Step 2: Use intercept form for a line with equal intercepts.

From the context, equation of line making intercepts $a$ and $b$ is $\frac{x}{a} + \frac{y}{b} = 1$.

For equal intercepts, let $a = b$. Then:

$$\frac{x}{a} + \frac{y}{a} = 1$$

or $x + y = a$

Step 3: Find the value of $a$ using the intersection point.

Since the line passes through $\left(\frac{1}{13}, \frac{5}{13}\right)$:

$$\frac{1}{13} + \frac{5}{13} = a$$ $$a = \frac{6}{13}$$

Step 4: Write the final equation.

$$x + y = \frac{6}{13}$$

or $13x + 13y - 6 = 0$