Let's solve this step by step.

Step 1: Vector Calculation

Start by finding the direction vector from the point $(3, 4, 0)$ to the point $(-1, -2, 2)$: $\vec{d} = (-1 - 3, -2 - 4, 2 - 0) = (-4, -6, 2)$

Step 2: Parametric Equation

The parametric equation for the line in the direction of this vector is: $(x, y, z) = (3, 4, 0) + t(-4, -6, 2)$ $(x, y, z) = (3 - 4t, 4 - 6t, 2t)$

Step 3: Sphere Equation

Assuming the sphere is centered at the origin with radius $R$, its equation is: $x^2 + y^2 + z^2 = R^2$

Step 4: Substituting and Solving

Substituting the parametric equations: $(3 - 4t)^2 + (4 - 6t)^2 + (2t)^2 = R^2$ Expanding: $(9 - 24t + 16t^2) + (16 - 48t + 36t^2) + (4t^2) = R^2$ $9 + 16 + 16t^2 + 36t^2 + 4t^2 - 24t - 48t = R^2$ $25 + 56t^2 - 72t = R^2$ $56t^2 - 72t + 25 - R^2 = 0$

Step 5: Discriminant Check

For the line to intersect the sphere, the discriminant of this quadratic equation must be non-negative: $\Delta = (-72)^2 - 4 \cdot 56 \cdot (25 - R^2)$ If $\Delta < 0$, then the point is not visible, and we need to check the other direction.

Would you like me to continue with detailed calculations or explain any step further?

Related Questions:

1. How do you determine if a point is visible from another point on a sphere?
2. What is the significance of the discriminant in determining the intersection?
3. How can parametric equations be used in other calculus problems?
4. What role does vector direction play in geometric visibility problems?
5. How would the problem change if the sphere's center was not at the origin?

Tip:

Always check the discriminant in quadratic equations involving distances or intersections—it reveals if a real solution exists.