Kirhoff’s voltage law is sometimes rendered as, “the sum of the voltage drops around a loop is zero.” For a bit of charge starting at point B and going around loop 1, the voltage rises through the power supply, drops through resistor R_{1}, and drops through resistor R_{3}, if the reference direction for R_{3} is seen as positive at point A and negative at point B. The voltage rise through the power supply V_{1} can be described as a negative voltage drop, so V_{1} appears as a negative term.

I note that there’s a discrepancy between the drawing and the equation you posted,

Bennington:

-V+Vr1+Vr3=0

Each resistor in the drawing has a polarity marker indicating the reference direction for the voltage across it. R_{3}’s polarity marker is at point B. Traveling aroung loop 1, the voltage across R_{3} comes up as a voltage rise rather than a voltage drop, so I’d expect to see its appearance in the equation as negative, like this:

-V_{1} + V_{R2} - V_{R3} = 0

The sign of V_{R3} works itself out by how the current is represented: The voltage across the resistor is the resistance multiplied by the current entering the polarity mark. In this case, that’s R_{3} * (I_{2} - I_{1}). I_{1} appears as a negative term because the reference direction for I_{1} shown in the drawing enters the other end of the resistor. That’s the opposite of the signs of the two loop currents shown in your post:

Vr3=(I1-I2)R3

so, the math in your post is consistent. It’s just not entirely consistent with the reference directions for voltage shown in the drawing.

*Edit: clarity*