Find Square of Distance from Point to Line in Three Dimensions

To find the square of the distance of the point $P(5, 6, 7)$ from the given line, we can determine the foot of the perpendicular from $P$ onto the line.

The equation of the line is: $\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4} = \lambda$

Any general point $Q$ on this line can be written in terms of the parameter $\lambda$ as: $Q = (2\lambda + 2, 3\lambda + 5, 4\lambda + 2)$

The direction ratios of the vector $\vec{PQ}$ joining the point $P(5, 6, 7)$ to $Q$ are: $\vec{PQ} = (2\lambda + 2 - 5)\hat{i} + (3\lambda + 5 - 6)\hat{j} + (4\lambda + 2 - 7)\hat{k}$ $\vec{PQ} = (2\lambda - 3)\hat{i} + (3\lambda - 1)\hat{j} + (4\lambda - 5)\hat{k}$

The direction vector of the line is $\vec{d} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

If $Q$ is the foot of the perpendicular from $P$ to the line, then $\vec{PQ}$ must be perpendicular to $\vec{d}$, which means their dot product is zero: $\vec{PQ} \cdot \vec{d} = 0$

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

Expanding and simplifying: $4\lambda - 6 + 9\lambda - 3 + 16\lambda - 20 = 0$ $29\lambda - 29 = 0$ $\lambda = 1$

Substituting $\lambda = 1$ into the expression for $\vec{PQ}$: $\vec{PQ} = (2(1) - 3)\hat{i} + (3(1) - 1)\hat{j} + (4(1) - 5)\hat{k} = -\hat{i} + 2\hat{j} - \hat{k}$

The square of the distance between point $P$ and the line is given by $PQ^2$: $PQ^2 = |\vec{PQ}|^2 = (-1)^2 + 2^2 + (-1)^2 = 1 + 4 + 1 = 6$

Thus, the square of the distance is equal to 6.