Find the exact coordinates of all possible points D on the line through A and B so that D is four times as far from A as it (D) is from B

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A(4, 7, -3) B(-3, 1, 2) AB <-7, -6, 5>

parametric equation for AB: x = 4 - 7t ; y = 7 - 6t ; z = -3 + 5t

I tried to use the distance formula where I set 4d (d being the distance of D to B) as the distance from D to A. I really don't know where to a) go from there because I got stuck or b) begin.

The question says "exact coordinates" so I would assume there are multiple coordinates that fit the criteria. I guess this means I'd have to make/find a general equation to find all the points but I don't know where to start on that either.

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Hint:

what is the point in $$ \begin{pmatrix} x\\y\\z \end{pmatrix}= \begin{pmatrix} 4\\7\\-3 \end{pmatrix} +t\begin{pmatrix} -7\\-6\\5 \end{pmatrix} $$ for $t=\frac{4}{5}$ ?

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There is a formula in coordinate geometry to find the coordinates of a point between 2 points $(x_1, y_1, z_1) and (x_2, y_2, z_3)$ in the ratio of m:n.

$$x = \frac{nx_1+mx_2}{m+n}$$ $$y = \frac{ny_1+my_2}{m+n}$$ $$z = \frac{nz_1+mz_2}{m+n}$$

There are 2 cases

Case 1 : D is between A and B Apply the above formula we have D equals $(\frac{-8}{5}, \frac{11}{5}, 1)$

Case 2 : D is closer to B on AB's extension, then treat B as the middle point and AB:BD = 3:1. We have $\frac{3x+4}{4} = -3$ $x=\frac{-16}{3}$

Similarly y=-1 and z=11/3.