Let N be the foot of perpendicular from the point $$P(1, -2, 3)$$ on the line passing through the points $$(4, 5, 8)$$ and $$(1, -7, 5)$$. Then the distance of N from the plane $$2x - 2y + z + 5 = 0$$ is

Line passes through $$A(4,5,8)$$ and $$B(1,-7,5)$$. Direction: $$\vec{d} = (3,12,3)$$ or simplified $$(1,4,1)$$.

Parametric form: $$(4+t, 5+4t, 8+t)$$ or $$(1+t, -7+4t, 5+t)$$.

Let's use: point on line = $$(4+t, 5+4t, 8+t)$$ with direction $$(1,4,1)$$.

Vector from P(1,-2,3) to point on line: $$(3+t, 7+4t, 5+t)$$

For foot of perpendicular, this must be perpendicular to direction $$(1,4,1)$$:

$$(3+t)(1) + (7+4t)(4) + (5+t)(1) = 0$$

$$3+t+28+16t+5+t = 0$$

$$36+18t = 0$$

$$t = -2$$

Foot N: $$(4-2, 5-8, 8-2) = (2, -3, 6)$$

Distance of N from plane $$2x - 2y + z + 5 = 0$$:

$$d = \frac{|2(2) - 2(-3) + 6 + 5|}{\sqrt{4+4+1}} = \frac{|4+6+6+5|}{3} = \frac{21}{3} = 7$$

This matches option 4: 7.