1. **State the problem:** We want to find the value of $b$ using the condition $f(2) = 3$ for the function $$f(x) = \frac{x^2 + 2x + b}{x - 1}.$$\n\n2. **Use the given point:** Substitute $x = 2$ and $f(2) = 3$ into the function:\n$$3 = \frac{2^2 + 2(2) + b}{2 - 1}.$$\n\n3. **Simplify:** Calculate numerator and denominator separately:\nNumerator: $$2^2 + 2(2) + b = 4 + 4 + b = 8 + b,$$\nDenominator: $$2 - 1 = 1.$$\nTherefore, $$3 = \frac{8 + b}{1} = 8 + b.$$\n\n4. **Solve for $b$:**\n$$3 = 8 + b \Rightarrow b = 3 - 8 = -5.$$\n\n5. **Write the function with $b$ found:**\n$$f(x) = \frac{x^2 + 2x - 5}{x - 1}.$$\n\n6. **Find columnar intersections:** Columnar intersections occur where the function intersects the vertical lines (vertical asymptotes or points of discontinuity). Since the denominator is $$x - 1,$$ the function is undefined at $$x=1.$$ Check if the function has a removable discontinuity or vertical asymptote at $$x=1$$ by factoring the numerator:\n$$x^2 + 2x - 5 = (x + 1)^2 - 6$$ (not factorable to include $$x-1$$). Therefore, no cancellation. Hence, there is a vertical asymptote (columnar intersection) at $$x=1.$$