Let P be the plane, which contains the line of intersection of the planes, $$x + y + z - 6 = 0$$ and $$2x + 3y + z + 5 = 0$$ and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to:

First, we collect the two given planes:

$$\Pi_1:\;x+y+z-6=0 \qquad\text{and}\qquad \Pi_2:\;2x+3y+z+5=0.$$

Every plane that contains their line of intersection can be written with the one-parameter family

$$\Pi:\;(\Pi_1)+\lambda(\Pi_2)=0.$$

That is,

$$\bigl(x+y+z-6\bigr)+\lambda\bigl(2x+3y+z+5\bigr)=0.$$

Expanding and collecting like terms we obtain

$$\;(1+2\lambda)\,x+(1+3\lambda)\,y+(1+\lambda)\,z+(-6+5\lambda)=0.$$

For the required plane to be perpendicular to the $$xy$$-plane, its normal vector must lie completely inside the $$xy$$-plane. The normal of the $$xy$$-plane (which is $$z=0$$) is $$\langle 0,0,1\rangle$$, so our plane’s normal must be orthogonal to this vector; therefore its $$z$$-component must vanish. Hence we set

$$1+\lambda=0\;\;\Longrightarrow\;\;\lambda=-1.$$

Substituting $$\lambda=-1$$ into the coefficients above gives

$$\begin{aligned} 1+2\lambda&=1+2(-1)=-1,\\[2pt] 1+3\lambda&=1+3(-1)=-2,\\[2pt] 1+\lambda&=0,\\[2pt] -6+5\lambda&=-6+5(-1)=-11. \end{aligned}$$

Thus the required plane $$P$$ is

$$-x-2y-11=0.$$

Multiplying by $$-1$$ for neatness, we may write

$$P:\;x+2y+11=0.$$

(Notice that if a point lies on the intersection line of $$\Pi_1$$ and $$\Pi_2$$, then from those two equations we have $$x+2y=-11$$, exactly the condition $$x+2y+11=0$$, verifying that the line indeed lies in $$P$$.)

Now we calculate the perpendicular distance from the point $$A(0,0,256)$$ to this plane. The distance formula from a point $$(x_0,y_0,z_0)$$ to a plane $$Ax+By+Cz+D=0$$ is

$$\text{Distance}=\dfrac{\lvert Ax_0+By_0+Cz_0+D\rvert}{\sqrt{A^2+B^2+C^2}}.$$

For the plane $$x+2y+11=0$$ we have $$A=1,\;B=2,\;C=0,\;D=11.$$ Substituting the coordinates of $$A$$ gives

Numerator:

$$\lvert 1\cdot0+2\cdot0+0\cdot256+11\rvert=\lvert 11\rvert=11.$$

Denominator:

$$\sqrt{1^2+2^2+0^2}=\sqrt{1+4}=\sqrt5.$$

Therefore

$$\text{Distance}=\dfrac{11}{\sqrt5}\;\text{units}.$$

Hence, the correct answer is Option C.