1. Set up the geometry

The points $P$ and $Q$ lie on the line $3x+4y=12.$ Also, $O=(0,0)$ and \angle POQ=90^\circ.

Since $\triangle OPQ$ is isosceles and the right angle is at $O$, the only possibility is $OP=OQ.$ So $\triangle OPQ$ is a right isosceles triangle with legs $OP$ and $OQ$.

2. Use the chord property

Because $P,Q$ lie on the line $3x+4y=12$, the segment $PQ$ is a chord of the circle centered at $O$ with radius $r=OP=OQ.$

Since $\angle POQ=90^\circ$, the chord length is $PQ = 2r\sin\frac{90^\circ}{2}=2r\sin45^\circ=r\sqrt{2}.$

Hence $PQ^2=2r^2.$

Therefore, $l=OP^2+PQ^2+QO^2=r^2+2r^2+r^2=4r^2.$

So we only need $r^2$.

3. Condition for the line to cut a chord subtending $90^\circ$ at the center

For a circle centered at the origin with radius $r$, if a chord subtends angle $90^\circ$ at the center, then the perpendicular distance $d$ from the center to the chord satisfies $d=r\cos45^\circ=\frac{r}{\sqrt2}.$ So $r=d\sqrt2.$

Now find the distance from the origin to the line $3x+4y=12$: $d=\frac{| -12 |}{\sqrt{3^2+4^2}}=\frac{12}{5}.$

Thus $r=\frac{12}{5}\sqrt2,$ so $r^2=\frac{288}{25}.$

Hence $l=4r^2=4\cdot \frac{288}{25}=\frac{1152}{25}=46.08.$

4. Greatest integer less than or equal to $l$

$\lfloor l\rfloor=\lfloor 46.08\rfloor=46.$

5. Check options

• A: $42$ ❌
• B: $46$ ✅
• C: $48$ ❌
• D: $44$ ❌

So the correct option is B.