If a point $$R(4, y, z)$$ lies on the line segment joining the points $$P(2, -3, 4)$$ and $$Q(8, 0, 10)$$, then the distance of R from the origin is:

We are given the points $$P(2,-3,4)$$ and $$Q(8,0,10)$$, and we know that the unknown point $$R(4,y,z)$$ lies somewhere on the line segment joining these two points.

Any point on the segment $$PQ$$ can be written with the help of the section (or “internal‐division”) formula. In its vector (or coordinate‐wise) form, the formula states:

$$\bigl(x,\;y,\;z\bigr)\;=\;\Bigl(\,x_1+t\,(x_2-x_1),\;y_1+t\,(y_2-y_1),\;z_1+t\,(z_2-z_1)\Bigr),$$

where

$$P(x_1,y_1,z_1)=P(2,-3,4),\qquad Q(x_2,y_2,z_2)=Q(8,0,10),$$

and the real parameter $$t$$ satisfies $$0\le t\le 1$$ when the point is actually between $$P$$ and $$Q$$.

Applying this to our data, we write the coordinates of a general point on $$PQ$$ as

$$\bigl(x,\;y,\;z\bigr) =\Bigl(2+t(8-2),\; -3+t(0-(-3)),\; 4+t(10-4)\Bigr) =\bigl(2+6t,\;-3+3t,\;4+6t\bigr).$$

Because the point $$R$$ has first coordinate $$x=4$$, we equate

$$2+6t = 4.$$

Solving this simple linear equation, we have

$$6t = 4-2 = 2 \;\;\Longrightarrow\;\; t=\dfrac{2}{6}=\dfrac13.$$

Now that we know $$t=\dfrac13$$, we substitute this value into the expressions for $$y$$ and $$z$$:

For the $$y$$‐coordinate, $$y = -3 + 3t = -3 + 3\left(\dfrac13\right) = -3 + 1 = -2.$$

For the $$z$$‐coordinate, $$z = 4 + 6t = 4 + 6\left(\dfrac13\right) = 4 + 2 = 6.$$

Hence the concrete coordinates of the point $$R$$ are

$$R(4,\,-2,\;6).$$

Next, we calculate the distance of this point from the origin $$O(0,0,0)$$. The distance formula in three dimensions is

$$\text{Distance} = \sqrt{x^2 + y^2 + z^2}.$$

Substituting $$x=4,\;y=-2,\;z=6$$, we get

$$\sqrt{\,4^{2} + (-2)^{2} + 6^{2}} = \sqrt{\,16 + 4 + 36} = \sqrt{56} = \sqrt{4\cdot14} = 2\sqrt{14}.$$

This numerical value matches option D in the list provided.

Hence, the correct answer is Option D.