The coordinates of the foot perpendicular from the point $$(1, 0, 0)$$ to the line $$\dfrac{x-1}{2} = \dfrac{y+1}{-3} = \dfrac{z+10}{8}$$ are

Let the given point be $$P(1,0,0)$$ and let the required foot of the perpendicular be $$Q(x,y,z)$$ on the line

$$\dfrac{x-1}{2}=\dfrac{y+1}{-3}=\dfrac{z+10}{8}=t$$

Writing the parametric form, any point on the line can be expressed as
$$Q(t):\; x=1+2t,\; y=-1-3t,\; z=-10+8t$$

The direction vector of the line is
$$\mathbf{d}= \langle 2,\,-3,\,8 \rangle$$

The vector $$\overrightarrow{PQ}$$ from $$P$$ to an arbitrary point $$Q(t)$$ on the line is
$$\overrightarrow{PQ}= \langle 1+2t-1,\; -1-3t-0,\; -10+8t-0 \rangle =\langle 2t,\; -1-3t,\; -10+8t \rangle$$

For $$Q$$ to be the foot of the perpendicular, $$\overrightarrow{PQ}$$ must be orthogonal to the line’s direction vector $$\mathbf{d}$$, i.e.
$$\overrightarrow{PQ}\cdot \mathbf{d}=0$$

Compute the dot product:
$$\langle 2t,\,-1-3t,\,-10+8t \rangle \cdot \langle 2,\,-3,\,8 \rangle = 2t\cdot2 + (-1-3t)(-3) + (-10+8t)\cdot8$$ $$= 4t + 3 + 9t - 80 + 64t$$ $$= (4+9+64)t + (3-80)$$ $$= 77t - 77$$

Setting this equal to zero:
$$77t - 77 = 0 \;\Longrightarrow\; t = 1$$

Substitute $$t = 1$$ back into the parametric equations:

$$x = 1 + 2(1) = 3$$
$$y = -1 - 3(1) = -4$$
$$z = -10 + 8(1) = -2$$

Therefore, the coordinates of the foot of the perpendicular are $$(3,\,-4,\,-2)$$.

Option D which is: $$(3, -4, -2)$$