We are given the following setup:

• You are standing at point $(5,6,0)$ and looking in a direction.
• The equation of the sphere is $x^2 + y^2 + z^2 = 1$.
• Two directions are given: either towards point $(-1,-2,2)$ or $(-1,-2,1)$.

The goal is to find the intersection point(s) where the direction from your position touches the surface of the sphere.

Step 1: Parametrize the direction vector

The direction vector from point $(5,6,0)$ to either of the points $(-1,-2,2)$ or $(-1,-2,1)$ is given by:

• For $(-1,-2,2)$, the vector is: $\vec{v_1} = (-1 - 5, -2 - 6, 2 - 0) = (-6, -8, 2)$
• For $(-1,-2,1)$, the vector is: $\vec{v_2} = (-1 - 5, -2 - 6, 1 - 0) = (-6, -8, 1)$

Thus, the parametric equation of the line is: $\vec{r}(t) = \vec{r_0} + t \cdot \vec{v}$ where $\vec{r_0} = (5,6,0)$ is the starting point and $\vec{v}$ is the direction vector.

Step 2: Find the intersection of the line and the sphere

The equation of the sphere is: $x^2 + y^2 + z^2 = 1$ Substitute the parametric coordinates of the line into this equation: $(x(t))^2 + (y(t))^2 + (z(t))^2 = 1$ We will handle this process for each direction vector:

Case 1: Direction vector $\vec{v_1} = (-6, -8, 2)$

The parametric coordinates of the line in this case are: $x(t) = 5 - 6t, \quad y(t) = 6 - 8t, \quad z(t) = 0 + 2t$ Substitute these into the sphere equation: $(5 - 6t)^2 + (6 - 8t)^2 + (2t)^2 = 1$ Simplify and solve for $t$.

Case 2: Direction vector $\vec{v_2} = (-6, -8, 1)$

The parametric coordinates of the line in this case are: $x(t) = 5 - 6t, \quad y(t) = 6 - 8t, \quad z(t) = 0 + t$ Substitute these into the sphere equation: $(5 - 6t)^2 + (6 - 8t)^2 + (t)^2 = 1$ Simplify and solve for $t$.

Let me calculate the solutions.