We need to check the quadrant which may contain the point $$(x, y)$$ that satisfies the inequality $$x+3 y<1$$

We can check the given inequality for each of the four quadrants one by one.

(Note: - As we are looking for the possibility of the inequality $x$+3 y<1$$ to be true for different quadrants, we will try for those points in different quadrants for which the inequality holds true)

Quadrant I - If the point $$(x, y)=\left(\frac{1}{3}, \frac{1}{6}\right)$$, it satisfies the inequality $$x+3 y=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}<1$$

Quadrant II - If we consider the point $$(x, y)=(-4,1)$$, it satisfies the inequality $$x+3 y--4+3 \times 1--4+3--1<1$$

Quadrant III - If we consider $$(x, y)=(-3,-2)$$, it satisfies the inequality $$x+3 y=-2+3 x-2=-2-6=-8<1$$

Quadrant IV - If we consider $$(x, y)=(1,-2)$$, it satisfies the inequality $$x+3 y=1+3 x-2=1-6=-5<1$$

Hence all four quadrants may contain the point $$(x, y)$$, so the answer is ( $$E$$ ).