Four equal masses, $$m$$ each are placed at the corners of a square of length $$l$$ as shown in the figure. The moment of inertia of the system about an axis passing through $$A$$ and parallel to $$DB$$ would be:

Let us place the square with vertices $$A$$, $$B$$, $$C$$, $$D$$ at coordinates $$A = (0, 0)$$, $$B = (l, 0)$$, $$C = (l, l)$$, and $$D = (0, l)$$. The diagonal $$DB$$ goes from $$(0, l)$$ to $$(l, 0)$$, so its direction vector is $$(1, -1)$$ (or unit vector $$\frac{1}{\sqrt{2}}(1, -1)$$).

The axis passes through $$A = (0, 0)$$ and is parallel to $$DB$$. The line through the origin with direction $$(1, -1)$$ has equation $$x + y = 0$$, so the perpendicular distance of a point $$(x, y)$$ from this axis is $$d = \frac{|x + y|}{\sqrt{2}}$$.

For mass at $$A = (0,0)$$: $$d_A = 0$$, so $$d_A^2 = 0$$.

For mass at $$B = (l, 0)$$: $$d_B = \frac{l}{\sqrt{2}}$$, so $$d_B^2 = \frac{l^2}{2}$$.

For mass at $$C = (l, l)$$: $$d_C = \frac{2l}{\sqrt{2}} = l\sqrt{2}$$, so $$d_C^2 = 2l^2$$.

For mass at $$D = (0, l)$$: $$d_D = \frac{l}{\sqrt{2}}$$, so $$d_D^2 = \frac{l^2}{2}$$.

The total moment of inertia is $$I = m(d_A^2 + d_B^2 + d_C^2 + d_D^2) = m\left(0 + \frac{l^2}{2} + 2l^2 + \frac{l^2}{2}\right) = m \times 3l^2 = 3ml^2$$.