Let $$P$$ be the point $$(10, -2, -1)$$ and $$Q$$ be the foot of the perpendicular drawn from the point $$R(1, 7, 6)$$ on the line passing through the points $$(2, -5, 11)$$ and $$(-6, 7, -5)$$. Then the length of the line segment $$PQ$$ is equal to ________

The line passes through $$A(2, -5, 11)$$ and $$B(-6, 7, -5)$$.

Direction vector ($$\vec{v}$$): $$\vec{AB} = (-6-2, 7-(-5), -5-11) = (-8, 12, -16)$$.

Simplified direction vector: (2, -3, 4) (dividing by -4).

Equation of the line: $$\vec{r} = (2, -5, 11) + \lambda(2, -3, 4)$$

Q lies on the line, so $$Q = (2 + 2\lambda, -5 - 3\lambda, 11 + 4\lambda)$$

The vector $$\vec{RQ}$$ is perpendicular to the line’s direction (2, -3, 4).

$$\vec{RQ} = (2\lambda + 1, -3\lambda - 12, 4\lambda + 5)$$

$$\text{Dot product: } 2(2\lambda + 1) - 3(-3\lambda - 12) + 4(4\lambda + 5) = 0$$

$$4\lambda + 2 + 9\lambda + 36 + 16\lambda + 20 = 0 \implies 29\lambda + 58 = 0 \implies \lambda = -2$$

Plugging $$\lambda = -2$$ into the coordinates of Q:

$$Q = (2 - 4, -5 + 6, 11 - 8) = (-2, 1, 3)$$

$$P(10, -2, -1), Q(-2, 1, 3)$$

$$PQ = \sqrt{(-2-10)^2 + (1-(-2))^2 + (3-(-1))^2}$$

$$PQ = \sqrt{(-12)^2 + 3^2 + 4^2} = \sqrt{144 + 9 + 16} = \sqrt{169} = \mathbf{13}$$