If $$a \neq 0$$ and the line $$2bx + 3cy + 4d = 0$$ passes through the points of intersection of the parabolas $$y^2 = 4ax$$ and $$x^2 = 4ay$$, then

The equations of the two given parabolas are:

$$ y^2 = 4ax $$

$$ x^2 = 4ay $$

To find their points of intersection, express $$ y $$ from the second equation:

$$ y = \frac{x^2}{4a} $$

Substitute this expression for $$ y $$ into the first parabola equation:

$$ \left(\frac{x^2}{4a}\right)^2 = 4ax $$

$$ \frac{x^4}{16a^2} = 4ax $$

Multiply both sides by the denominator to clear the fraction:

$$ x^4 = 64a^3x $$

Move all terms to one side to solve the polynomial equation:

$$ x^4 - 64a^3x = 0 $$

$$ x(x^3 - 64a^3) = 0 $$

This equation gives two possible solutions for $$ x $$:

$$ x = 0 $$

$$ x^3 = 64a^3 $$

$$ x = 4a $$

Now substitute these $$ x $$ values back to find the corresponding $$ y $$ coordinates.

Case 1: When $$ x = 0 $$, the value of $$ y $$ is:

$$ y = \frac{0^2}{4a} = 0 $$

So, the first intersection point is the origin:

$$ (0, 0) $$

Case 2: When $$ x = 4a $$, the value of $$ y $$ is:

$$ y = \frac{(4a)^2}{4a} = 4a $$

So, the second intersection point is:

$$ (4a, 4a) $$

The problem states that the straight line passes through the points of intersection:

$$ 2bx + 3cy + 4d = 0 $$

Since the line passes through the first point $$ (0, 0) $$, substitute these coordinates into the line equation:

$$ 2b(0) + 3c(0) + 4d = 0 $$

$$ 4d = 0 $$

$$ d = 0 $$

Since the line also passes through the second point $$ (4a, 4a) $$, substitute these coordinates and the value of $$ d $$ into the line equation:

$$ 2b(4a) + 3c(4a) + 4(0) = 0 $$

$$ 8ab + 12ac = 0 $$

Factor out the common multiplier from the terms:

$$ 4a(2b + 3c) = 0 $$

Since the problem explicitly states that $$ a \neq 0 $$, we can safely divide both sides by $$ 4a $$:

$$ 2b + 3c = 0 $$

Final Answer:

from options , $$d^2 + (2b + 3c)^2 = 0$$ is correct