The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes $$3x + y - 2z = 5$$ and $$2x - 5y - z = 7$$, is

We need to find the equation of the plane passing through $$(1, 2, -3)$$ and perpendicular to both the planes $$3x + y - 2z = 5$$ and $$2x - 5y - z = 7$$.

The normal vectors of the two given planes are $$\vec{n_1} = (3, 1, -2)$$ and $$\vec{n_2} = (2, -5, -1)$$.

Since the required plane is perpendicular to both given planes, its normal vector is parallel to $$\vec{n_1} \times \vec{n_2}$$.

We compute the cross product:

$$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 2 & -5 & -1 \end{vmatrix}$$

$$= \hat{i}[(1)(-1) - (-2)(-5)] - \hat{j}[(3)(-1) - (-2)(2)] + \hat{k}[(3)(-5) - (1)(2)]$$

$$= \hat{i}[-1 - 10] - \hat{j}[-3 + 4] + \hat{k}[-15 - 2]$$

$$= -11\hat{i} - \hat{j} - 17\hat{k}$$

So the normal direction is $$(11, 1, 17)$$ (taking the negative).

The equation of the plane passing through $$(1, 2, -3)$$ with normal $$(11, 1, 17)$$ is $$11(x - 1) + 1(y - 2) + 17(z + 3) = 0$$.

Expanding, $$11x - 11 + y - 2 + 17z + 51 = 0$$, which gives $$11x + y + 17z + 38 = 0$$.

Hence, the correct answer is Option A.