The square of the distance of the point P(5, 6, 7) from the line...

Let the foot of the perpendicular from $P(5, 6, 7)$ to the given line be $Q$.

Any point on the line $\dfrac{x-2}{2} = \dfrac{y-5}{3} = \dfrac{z-2}{4} = \lambda$ is $Q(2\lambda + 2, 3\lambda + 5, 4\lambda + 2)$.

The direction ratios of $PQ$ are $(2\lambda - 3, 3\lambda - 1, 4\lambda - 5)$.

Since $PQ$ is perpendicular to the line, the dot product of their direction ratios is zero:

$2(2\lambda - 3) + 3(3\lambda - 1) + 4(4\lambda - 5) = 0$

$4\lambda - 6 + 9\lambda - 3 + 16\lambda - 20 = 0$

$29\lambda - 29 = 0 \Rightarrow \lambda = 1$

Substituting $\lambda = 1$, the coordinates of $Q$ are $(4, 8, 6)$.

The square of the distance $PQ^2$ is $(5 - 4)^2 + (6 - 8)^2 + (7 - 6)^2 = 1 + 4 + 1 = 6$.

Answer: $6$