1. Find equations of the sides of triangle $ABC$

Given: $A(-1,3),\quad B(-2,2),\quad C(3,-1)$

Side $AB$

Slope of $AB$: $m_{AB}=\frac{2-3}{-2-(-1)}=\frac{-1}{-1}=1$ So equation through $A(-1,3)$ is $y-3=1(x+1)$ $y=x+4$ or $x-y+4=0$

Side $BC$

Slope of $BC$: $m_{BC}=\frac{-1-2}{3-(-2)}=\frac{-3}{5}$ Equation through $B(-2,2)$: $y-2=-\frac35(x+2)$ $5y-10=-3x-6$ $3x+5y-4=0$

Side $CA$

Slope of $CA$: $m_{CA}=\frac{3-(-1)}{-1-3}=\frac{4}{-4}=-1$ Equation through $A(-1,3)$: $y-3=-(x+1)$ $x+y-2=0$

So the three sides are: $AB: x-y+4=0$ $BC: 3x+5y-4=0$ $CA: x+y-2=0$

2. Shift each side inward by 1 unit

For a line $ax+by+c=0$ shifting parallel by distance $1$ changes constant term by $\pm \sqrt{a^2+b^2}$ So the shifted line is $ax+by+c' =0$ with $|c'-c|=\sqrt{a^2+b^2}$ We must choose the sign so that the new line lies inside the triangle.

To determine the inward side, use any interior point of triangle $ABC$.

Take centroid: $G\left(\frac{-1-2+3}{3},\frac{3+2-1}{3}\right)=\left(0,\frac43\right)$ This lies inside the triangle.

Shift of $AB: x-y+4=0$

At centroid: $0-\frac43+4=\frac83>0$ Interior is on the side where $x-y+4>0$. To move inward by 1 unit, line should move toward centroid, so constant must decrease: $x-y+4-\sqrt{1^2+(-1)^2}=0$ $x-y+4-\sqrt2=0$

Shift of $BC: 3x+5y-4=0$

At centroid: $3(0)+5\cdot\frac43-4=\frac{20}{3}-4=\frac83>0$ Interior is where $3x+5y-4>0$. So inward shift: $3x+5y-4-\sqrt{3^2+5^2}=0$ $3x+5y-4-\sqrt{34}=0$

Shift of $CA: x+y-2=0$

At centroid: $0+\frac43-2=-\frac23<0$ Interior is where $x+y-2<0$. To move inward, constant must increase: $x+y-2+\sqrt{1^2+1^2}=0$ $x+y-2+\sqrt2=0$ $x+y-(2-\sqrt2)=0$

3. Identify which new side is nearest to origin

The three shifted sides are: $L_1: x-y+4-\sqrt2=0$ $L_2: 3x+5y-4-\sqrt{34}=0$ $L_3: x+y-(2-\sqrt2)=0$

Distance of origin from line $ax+by+c=0$ is $d=\frac{|c|}{\sqrt{a^2+b^2}}$

For $L_1$

$d_1=\frac{|4-\sqrt2|}{\sqrt2}=\frac{4-\sqrt2}{\sqrt2}=2\sqrt2-1$

For $L_2$

$d_2=\frac{|-(4+\sqrt{34})|}{\sqrt{34}}=\frac{4+\sqrt{34}}{\sqrt{34}}>1$

For $L_3$

$d_3=\frac{|-(2-\sqrt2)|}{\sqrt2}=\frac{2-\sqrt2}{\sqrt2}=\sqrt2-1$

Clearly, $\sqrt2-1 < 2\sqrt2-1$ and also $d_2$ is much larger. Hence the nearest side to the origin is $x+y-(2-\sqrt2)=0$

4. Match with options

This is Option B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.