1. Set up coordinates

Let the square have side $l$, with coordinates:

$A(0,0),\quad B(l,0),\quad D(0,l),\quad C(l,l)$

The diagonal $BD$ has slope

$\frac{l-0}{0-l}=-1$

So any line parallel to $BD$ also has slope $-1$.

The required axis passes through $A(0,0)$ and is parallel to $BD$, so its equation is

$y=-x$

or

$x+y=0$

2. Use perpendicular distance formula

Moment of inertia about a line is

$I=\sum m r_i^2$

where $r_i$ is the perpendicular distance of each mass from the axis.

For line $x+y=0$, perpendicular distance of point $(x_1,y_1)$ is

$r=\frac{|x_1+y_1|}{\sqrt{1^2+1^2}}=\frac{|x_1+y_1|}{\sqrt2}$

3. Find distances of all four masses from the axis

• At $A(0,0)$:

$r_A=\frac{|0+0|}{\sqrt2}=0$

• At $B(l,0)$:

$r_B=\frac{|l+0|}{\sqrt2}=\frac{l}{\sqrt2}$

• At $D(0,l)$:

$r_D=\frac{|0+l|}{\sqrt2}=\frac{l}{\sqrt2}$

• At $C(l,l)$:

$r_C=\frac{|l+l|}{\sqrt2}=\frac{2l}{\sqrt2}=\sqrt2\,l$

4. Compute moment of inertia

$I=m r_A^2+m r_B^2+m r_D^2+m r_C^2$

$I=m\left(0^2+\left(\frac{l}{\sqrt2}\right)^2+\left(\frac{l}{\sqrt2}\right)^2+(\sqrt2 l)^2\right)$

$I=m\left(0+\frac{l^2}{2}+\frac{l^2}{2}+2l^2\right)$

$I=m(3l^2)$

$\boxed{I=3ml^2}$

5. Match with options

This corresponds to:

$\boxed{\text{Option C}}$

6. Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So the derived answer agrees with the stored correct answer.