Question:

Step 1: Understand the condition for the locus equation to hold for all values of parameters. The equation $\alpha(2x - y + 1) + \beta(3x - y) + \gamma(2x + y - 5) = 0$ must be satisfied for all real values of $\alpha$, $\beta$, and $\gamma$. This is only possible if each coefficient (the expression multiplying each parameter) equals zero independently. Step 2: Set up the system of equations from the coefficients. For the equation to hold for all $\alpha, \beta, \gamma \in \mathbb{R}$, we require: $$2x - y + 1 = 0 \quad \text{...(1)}$$ $$3x - y = 0 \quad \text{...(2)}$$ $$2x + y - 5 = 0 \quad \text{...(3)}$$ Step 3: Solve for x by subtracting equations (1) and (2). Subtracting equation (2) from equation (1): $$(2x - y + 1) - (3x - y) = 0$$ $$2x - y + 1 - 3x + y = 0$$ $$-x + 1 = 0$$ $$x = 1$$ Step 4: Solve for y using equation (2). Substituting $x = 1$ into equation (2): $$3(1) - y = 0$$ $$y = 3$$ Step 5: Verify the solution satisfies equation (3). Checking with equation (3): $$2(1) + 3 - 5 = 2 + 3 - 5 = 0 \quad \checkmark$$ The solution is consistent with all three equations. Step 6: Identify the locus f(x, y). The locus $f(x, y)$ is the single point $P(1, 3)$. Step 7: Calculate the distance from the point to the given line. The distance from a point $(x_0, y_0)$ to a line $ax + by + c = 0$ is given by: $$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$ For the point $(1, 3)$ and the line $3x - 4y + 19 = 0$: $$d = \frac{|3(1) - 4(3) + 19|}{\sqrt{3^2 + (-4)^2}}$$ $$d = \frac{|3 - 12 + 19|}{\sqrt{9 + 16}}$$ $$d = \frac{|10|}{\sqrt{25}}$$ $$d = \frac{10}{5} = 2$$ **Final Answer:** The least distance between the curve $f(x, y)$ (which is the point $(1, 3)$) and the straight line $3x - 4y + 19 = 0$ is $\boxed{2}$. This corresponds to **Option 1**.