If one of the lines given by $$6x^2 - xy + 4cy^2 = 0$$ is $$3x + 4y = 0$$, then $$c$$ equals

Let the given pair of straight lines be represented by the equation:

$$ 6x^2 - xy + 4cy^2 = 0 $$

One of the lines given by this equation is:

$$ 3x + 4y = 0 $$

From this line equation, we can express $$ x $$ in terms of $$ y $$:

$$ 3x = -4y $$

$$ x = -\frac{4y}{3} $$

Since this line is part of the pair of lines, any point on it must satisfy the joint equation. Substitute this expression for $$ x $$ into the joint equation:

$$ 6\left(-\frac{4y}{3}\right)^2 - \left(-\frac{4y}{3}\right)y + 4cy^2 = 0 $$

Expand the squared term and simplify the multiplication:

$$ 6\left(\frac{16y^2}{9}\right) + \frac{4y^2}{3} + 4cy^2 = 0 $$

Reduce the fraction in the first term by dividing $$ 6 $$ and $$ 9 $$ by their common factor $$ 3 $$:

$$ 2\left(\frac{16y^2}{3}\right) + \frac{4y^2}{3} + 4cy^2 = 0 $$

$$ \frac{32y^2}{3} + \frac{4y^2}{3} + 4cy^2 = 0 $$

Combine the first two terms since they have a common denominator:

$$ \frac{36y^2}{3} + 4cy^2 = 0 $$

$$ 12y^2 + 4cy^2 = 0 $$

Factor out the common term $$ y^2 $$ from the equation:

$$ y^2(12 + 4c) = 0 $$

Since this equation holds true for all points along the line where $$ y \neq 0 $$, the expression inside the parentheses must be equal to $$ 0 $$:

$$ 12 + 4c = 0 $$

Solve for $$ c $$ by isolating the variable:

$$ 4c = -12 $$

$$ c = -3 $$

Final Answer:

The value of $$ c $$ is $$ -3 $$.