The equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $3 x+4 y+5 z=2$ and perpendicular to the XY- plane is

A plane passing through the line of intersection of $x+y+z-1=0$ and $3x+4y+5z-2=0$ has the form $(x+y+z-1) + \lambda(3x+4y+5z-2) = 0$.

Simplifying gives $(1+3\lambda)x + (1+4\lambda)y + (1+5\lambda)z - (1+2\lambda) = 0$ with normal vector $\vec{n} = (1+3\lambda)\hat{i} + (1+4\lambda)\hat{j} + (1+5\lambda)\hat{k}$.

Since this plane is perpendicular to the XY-plane $z=0$ with normal $\hat{k}$, their dot product must vanish: $\vec{n} \cdot \hat{k} = 1+5\lambda = 0$.

Solving yields $\lambda = -\frac{1}{5}$.

Substituting gives $\frac{2}{5}x + \frac{1}{5}y - \frac{3}{5} = 0$, which after multiplying by 5 becomes $2x + y - 3 = 0$.

The final answer is $\boxed{\text{A}}$.