The position vector of the point $A$ is $2\vec{i} - \vec{k}$ and the equation of the the line is:
$$\vec{r} = (-7, 15, -5) + \lambda (3, -7, 4) \ . $$
Find the position vectors of points $B$ and $C$, both lying on the line, such that the length $\overline{AB} = \overline{AC} = 10$.
So this is what I've done so far:
I found the foot of the perpendicular from point $A$ to the line and labeled that $M$. Using the fact that point $M$ lies on the line and that it is perpendicular to the director vector of the line, I found the coordinate of point $M$ (-1, 1, 3). Now that I have the position vector of point $M$ and the magnitude of $\overline{AC}$, I can find the magnitude of $\overline{MC}$ using the Pythagoras Theorem but I don't know how to continue from there and or if I am at all on the right path.
You can solve it the way you are working, but it seems to me that it's going to give you a bit more work that just solving: $$ \overline{AR}^2 = 10^2$$ with $R=( -7 + 3 \lambda, 15 -7 \lambda, -5 + 4 \lambda )$.
It will give a quadratic polynomial for $\lambda$, hence the two solutions.
Can you solve this?
Edit:
Since $A = (2, 0, -1)$, then the square of the distance between $A$ and the generic point of the straight line $r$ is:
$$ \overline{AR}^2 = (-7 + 3 \lambda - 2)^2 + (15 -7 \lambda - 0)^2 + (-5 + 4 \lambda - 1)^2 $$
Since we want $\overline{AR}^2 = 10^2$, we get $$ (-7 + 3 \lambda - 2)^2 + (15 -7 \lambda - 0)^2 + (-5 + 4 \lambda - 1)^2 = 100 $$
Which gives: $$ (3 \lambda - 9)^2 + (15 -7 \lambda)^2 + (4 \lambda - 6)^2 = 100 $$ $$ 9 \lambda^2 - 54 \lambda + 81 + 225 -210 \lambda + 49 \lambda^2 + 16 \lambda^2 -24 \lambda + 36 = 100 $$
Now, you just need to solve the quadratic polynomial in $\lambda$.