The square of the distance of the point $$P(5, 6, 7)$$ from the line $$\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}$$ is equal to:

To find the square of the distance of point $$P(5, 6, 7)$$ from the line $$\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4}$$, we need to find the foot of the perpendicular from point $$P$$ to the line.

Set the equation of the line equal to a parameter $$\lambda$$:

$$\frac{x-2}{2} = \frac{y-5}{3} = \frac{z-2}{4} = \lambda$$

Any general point $$M$$ on this line can be expressed as:

• $$x = 2\lambda + 2$$
• $$y = 3\lambda + 5$$
• $$z = 4\lambda + 2$$
• $$DR_x = (2\lambda + 2) - 5 = 2\lambda - 3$$
• $$DR_y = (3\lambda + 5) - 6 = 3\lambda - 1$$
• $$DR_z = (4\lambda + 2) - 7 = 4\lambda - 5$$
• $$x = 2(1) + 2 = 4$$
• $$y = 3(1) + 5 = 8$$
• $$z = 4(1) + 2 = 6$$

So, $$M = (2\lambda + 2, 3\lambda + 5, 4\lambda + 2)$$.

The direction ratios of the line $$PM$$ (where $$M$$ is the foot of the perpendicular from $$P$$) are:

The line $$PM$$ is perpendicular to the given line. The direction ratios of the given line are $$(2, 3, 4)$$. The dot product of their direction ratios must be zero:

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

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

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

Substitute $$\lambda = 1$$ into the coordinates of $$M$$:

So, $$M = (4, 8, 6)$$.

Using the distance formula for $$P(5, 6, 7)$$ and $$M(4, 8, 6)$$:

$$PM^2 = (4 - 5)^2 + (8 - 6)^2 + (6 - 7)^2$$

$$PM^2 = (-1)^2 + (2)^2 + (-1)^2$$

$$PM^2 = 1 + 4 + 1 = 6$$

Final Answer: 6 (Option C)