The image contains a series of math problems related to vectors, planes, and spheres in 3D space. We need to solve problem IV, which involves finding the equation of the line (AB), proving that lines AB and AC are orthogonal, and finding the equation of a plane parallel to the line (AC). A, B, and C are given as $A(0,1,1)$, $B(-1,-2,2)$, and $C(1,2,5)$.

1. Problem Description

The image contains a series of math problems related to vectors, planes, and spheres in 3D space. We need to solve problem IV, which involves finding the equation of the line (AB), proving that lines AB and AC are orthogonal, and finding the equation of a plane parallel to the line (AC). A, B, and C are given as $A(0,1,1)$, $B(-1,-2,2)$, and $C(1,2,5)$.

2. Solution Steps

(a) Finding the equation of the line (AB).

The direction vector of line AB is given by $\vec{AB} = B - A = (-1-0, -2-1, 2-1) = (-1, -3, 1)$.

The parametric equation of the line passing through point $A(x_0, y_0, z_0)$ with direction vector $(a, b, c)$ is given by:

$x = x_0 + at$

$y = y_0 + bt$

$z = z_0 + ct$

Substituting the coordinates of point A (0, 1, 1) and the direction vector $\vec{AB} = (-1, -3, 1)$, we have:

$x = 0 - t = -t$

$y = 1 - 3t$

$z = 1 + t$

The symmetric form of the line equation is $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$.

So, $\frac{x - 0}{-1} = \frac{y - 1}{-3} = \frac{z - 1}{1}$, which simplifies to:

$\frac{x}{-1} = \frac{y - 1}{-3} = \frac{z - 1}{1}$.

(b) Proving that AB and AC are orthogonal.

We need to find the direction vector $\vec{AC} = C - A = (1-0, 2-1, 5-1) = (1, 1, 4)$.

Two lines are orthogonal if the dot product of their direction vectors is zero.

$\vec{AB} \cdot \vec{AC} = (-1)(1) + (-3)(1) + (1)(4) = -1 - 3 + 4 = 0$.

Since $\vec{AB} \cdot \vec{AC} = 0$, AB and AC are orthogonal.

(c) Finding the equation of a plane parallel to the line (AC).

Since the plane is parallel to the line AC, the direction vector $\vec{AC}$ will be parallel to the plane. Since the problem asks for *a* plane, not a specific one, it is not required to be through a specific point. Since $\vec{AB}$ and $\vec{AC}$ are perpendicular, we can find a normal vector to a plane that includes both lines by using the vector product: $\vec{n} = \vec{AB} \times \vec{AC}$.

$\vec{n} = \vec{AB} \times \vec{AC} = (-1, -3, 1) \times (1, 1, 4) = ((-3)(4) - (1)(1), (1)(1) - (-1)(4), (-1)(1) - (-3)(1)) = (-12-1, 1+4, -1+3) = (-13, 5, 2)$

The general equation of a plane is $ax + by + cz = d$, where $(a, b, c)$ is the normal vector to the plane. Since we have a normal vector (-13, 5, 2), the equation of the plane can be written as:

$-13x + 5y + 2z = d$. Since the plane only needs to be parallel to the line AC, we can set d=0:

$-13x + 5y + 2z = 0$ or $13x - 5y - 2z = 0$.

3. Final Answer

(a) Equation of line (AB): $\frac{x}{-1} = \frac{y - 1}{-3} = \frac{z - 1}{1}$.

(b) $\vec{AB} \cdot \vec{AC} = 0$, therefore AB and AC are orthogonal.

(c) Equation of a plane parallel to line (AC): $13x - 5y - 2z = 0$.