Question:

Step 1: Identify the direction vector of the line perpendicular to the given plane. The line passes through the origin and is perpendicular to the plane $2x - y - z = 4$. The normal vector to this plane is the direction vector of the line. Therefore, the direction vector is $(2, -1, -1)$. The parametric equation of the line is: $$\vec{r}(t) = t(2, -1, -1) = (2t, -t, -t)$$ Step 2: Substitute the parametric line equation into the plane equation. We need to find where this line intersects the plane $\vec{r} \cdot (3\hat{i} - 5\hat{j} + 2\hat{k}) = 6$, which can be written as: $$3x - 5y + 2z = 6$$ Substituting $(x, y, z) = (2t, -t, -t)$: $$3(2t) - 5(-t) + 2(-t) = 6$$ Step 3: Simplify and solve for the parameter $t$. Expanding the equation: $$6t + 5t - 2t = 6$$ $$9t = 6$$ $$t = \frac{2}{3}$$ Step 4: Find the coordinates of the intersection point. Substituting $t = \frac{2}{3}$ into the parametric equation: $$(x_0, y_0, z_0) = \left(2 \cdot \frac{2}{3}, -\frac{2}{3}, -\frac{2}{3}\right) = \left(\frac{4}{3}, -\frac{2}{3}, -\frac{2}{3}\right)$$ Step 5: Calculate the required expression $2x_0 - 3y_0 + z_0$. $$2x_0 - 3y_0 + z_0 = 2\left(\frac{4}{3}\right) - 3\left(-\frac{2}{3}\right) + \left(-\frac{2}{3}\right)$$ $$= \frac{8}{3} + \frac{6}{3} - \frac{2}{3}$$ $$= \frac{8 + 6 - 2}{3} = \frac{12}{3} = 4$$ Therefore, the value of $(2x_0 - 3y_0 + z_0) = 4$. The answer is **Option 4: 4**.