If one of the lines given by the equation $$2x^{2} + axy + 3y^{2} = 0$$ coincides with one of those given by $$2x^{2} + bxy — 3𝑦^{2} = 0$$, and the other lines represented by them are perpendicular then $$a^{2} + b^{2}$$ is .

The general equation of the line is given as $$y=mx+c$$. But, in the question, the equation of pairs of lines is given as $$2x^{2} + axy + 3y^{2} = 0$$ and $$2x^{2} + bxy — 3𝑦^{2} = 0$$, which has no constant term. The equation of a pair of lines is given as the product of the equations of both lines. If there are no constant terms, this means that the equation of both straight lines is of the form $$y=mx$$.

For $$2x^{2} + axy + 3y^{2} = 0\rightarrow1$$, let the equation of straight lines be $$y=mx$$ and $$y=nx$$.

=> $$y-mx=0$$ and $$y-nx=0$$. We will multiply the equations of two lines to get the equation of a pair of lines. 

=> $$(y-mx)(y-nx)=0$$

=> $$y^2-\left(m+n\right)xy+mnx^2=0\rightarrow2$$

Comparing eq. 1 and eq. 2, since both equations are of the same pair of straight lines.

=> $$\dfrac{2}{mn}=\dfrac{a}{-\left(m+n\right)}=\dfrac{3}{1}$$

=> $$mn=\dfrac{2}{3}\rightarrow3$$ and $$a=-3\left(m+n\right)\rightarrow4$$

For $$2x^{2} + axy + 3y^{2} = 0\rightarrow5$$, since one of the line is coinciding with the first pair of straight lines, the equation of that straight line will be $$y=mx$$ and and since the other line is perpendicular to the equation of the first pair of straight lines, it's equation will be $$y=-\dfrac{x}{n}$$. 

=> $$y-mx=0$$ and $$y+\dfrac{x}{n}=0$$. We will multiply the equations of two lines to get the equation of a pair of lines.

=> $$\left(y-mx\right)\left(y+\dfrac{x}{n}\right)=0$$

=> $$y^2+xy\left(\frac{1}{n}-m\right)-\frac{m}{n}x^2=0\rightarrow6$$

Comparing eq. 5 and eq. 6, since both equations are of the same pair of straight lines.

=> $$-\dfrac{\frac{m}{n}}{2}=\dfrac{\frac{1}{n}-m}{b}=\dfrac{1}{-3}$$

=> $$\dfrac{m}{n}=\dfrac{2}{3}\rightarrow7$$ and $$b=3\left(m-\dfrac{1}{n}\right)\rightarrow8$$

Multiplying eq. 3 and eq. 7 -

=> $$\dfrac{m}{n}\times mn=\dfrac{2}{3}\times\dfrac{2}{3}$$ => $$m^2=\dfrac{4}{9}$$ 

=> $$m=\pm\dfrac{2}{3}$$ and, thus $$n=\pm1$$

Using the values of m and n, we can calculate the value of a and b. It doesn't matter whether we take the positive values or the negative values of m and n, because we will get the same magnitudes of a and b, whether taking positive or negative, and since we need to calculate the value of $$a^2+b^2$$, thus we just need the magnitude as the sign will ultimately become positive only. Therefore, substituting the value of m and n in eq. 4 and eq. 8 - 

=> $$a=-3\left(m+n\right)=-3\left(\dfrac{2}{3}+1\right)=-3\left(\dfrac{5}{3}\right)=-5$$

=> $$b=3\left(m-\dfrac{1}{n}\right)=3\left(\dfrac{2}{3}-1\right)=3\left(-\dfrac{1}{3}\right)=-1$$

[If m = -2/3 and n = -1, then the value of a and b will be 5 and 1 respectively]

=> $$a^2+b^2=\left(-5\right)^2+\left(-1\right)^2=25+1=26$$