The equation of the plane passing through the point of intersection of the planes $2 x-y+z-3=0$ and $4 x-3 y+5 z+9=0$ and parallel to the lines $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$ is $\alpha x+\beta y+\gamma z+d=0$ Then $\alpha+\beta+\gamma+\mathrm{d}=$

Plane through intersection: $P_1 + \lambda P_2 = (2x - y + z - 3) + \lambda(4x - 3y + 5z + 9) = 0$

Rewriting: $(2 + 4\lambda)x + (-1 - 3\lambda)y + (1 + 5\lambda)z + (-3 + 9\lambda) = 0$

The plane is parallel to the line with direction vector $\vec{v} = 2\hat{i} + 4\hat{j} + 5\hat{k}$. The normal vector $\vec{n} = (2 + 4\lambda)\hat{i} + (-1 - 3\lambda)\hat{j} + (1 + 5\lambda)\hat{k}$ must be perpendicular to $\vec{v}$, so their dot product is zero:

$\vec{n} \cdot \vec{v} = (2 + 4\lambda)(2) + (-1 - 3\lambda)(4) + (1 + 5\lambda)(5) = 0$

Simplifying: $4 + 8\lambda - 4 - 12\lambda + 5 + 25\lambda = 0$
$5 + 21\lambda = 0$
$\lambda = -\frac{5}{21}$

Substitute back into the plane equation:
$(2 + 4(-\frac{5}{21}))x + (-1 - 3(-\frac{5}{21}))y + (1 + 5(-\frac{5}{21}))z + (-3 + 9(-\frac{5}{21})) = 0$
Simplifying: $\frac{22}{21}x - \frac{6}{21}y - \frac{4}{21}z - \frac{108}{21} = 0$
Multiply by 21: $22x - 6y - 4z - 108 = 0$

Simplify by dividing by 2: $11x - 3y - 2z - 54 = 0$

With $\alpha = 11$, $\beta = -3$, $\gamma = -2$, $d = -54$:
$\alpha + \beta + \gamma + d = 11 - 3 - 2 - 54 = -48$

The final answer is $\boxed{\text{B}}$