Given points $A(2, 0, 1)$, $B(0, 1, 3)$, and $C(0, 3, 2)$, we need to: a. Plot the points $A$, $B$, and $C$. b. Find the coordinates of vectors $\vec{BA}$, $\vec{BC}$, and $\vec{AC}$. c. Calculate $\vec{BA} \cdot \vec{BC}$ and determine if triangle $ABC$ is a right triangle. d. Find the general equation of a sphere $(S)$ centered at $A$ and passing through $B$. e. Find the equation of the plane defined by the vector $\vec{AB}$.

1. Problem Description

Given points $A(2, 0, 1)$, $B(0, 1, 3)$, and $C(0, 3, 2)$, we need to:

a. Plot the points $A$, $B$, and $C$.

b. Find the coordinates of vectors $\vec{BA}$, $\vec{BC}$, and $\vec{AC}$.

c. Calculate $\vec{BA} \cdot \vec{BC}$ and determine if triangle $ABC$ is a right triangle.

d. Find the general equation of a sphere $(S)$ centered at $A$ and passing through $B$.

e. Find the equation of the plane defined by the vector $\vec{AB}$.

2. Solution Steps

a. Plotting the points $A(2, 0, 1)$, $B(0, 1, 3)$, and $C(0, 3, 2)$ involves placing these points in a 3D coordinate system.

b. Finding the coordinates of the vectors:

$\vec{BA} = A - B = (2, 0, 1) - (0, 1, 3) = (2, -1, -2)$

$\vec{BC} = C - B = (0, 3, 2) - (0, 1, 3) = (0, 2, -1)$

$\vec{AC} = C - A = (0, 3, 2) - (2, 0, 1) = (-2, 3, 1)$

c. Calculating the dot product $\vec{BA} \cdot \vec{BC}$:

$\vec{BA} \cdot \vec{BC} = (2)(0) + (-1)(2) + (-2)(-1) = 0 - 2 + 2 = 0$

Since $\vec{BA} \cdot \vec{BC} = 0$, vectors $\vec{BA}$ and $\vec{BC}$ are orthogonal. Therefore, triangle $ABC$ is a right triangle with the right angle at vertex $B$.

d. Finding the equation of the sphere $(S)$ centered at $A$ and passing through $B$:

The general equation of a sphere with center $(a, b, c)$ and radius $r$ is:

$(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2$

Here, the center is $A(2, 0, 1)$. The radius $r$ is the distance between $A$ and $B$:

$r = \sqrt{(2 - 0)^2 + (0 - 1)^2 + (1 - 3)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

So, the equation of the sphere is:

$(x - 2)^2 + (y - 0)^2 + (z - 1)^2 = 3^2$

$(x - 2)^2 + y^2 + (z - 1)^2 = 9$

Expanding this, we get:

$x^2 - 4x + 4 + y^2 + z^2 - 2z + 1 = 9$

$x^2 + y^2 + z^2 - 4x - 2z + 5 - 9 = 0$

$x^2 + y^2 + z^2 - 4x - 2z - 4 = 0$

e. Finding the equation of the plane defined by the vector $\vec{AB}$:

First, find the vector $\vec{AB} = B - A = (0, 1, 3) - (2, 0, 1) = (-2, 1, 2)$.

However, a single vector doesn't define a plane. We need a point and a normal vector to the plane, or three points. Since the problem is ambiguous, I cannot determine the plane. Assuming the problem is asking for the equation of a plane normal to the vector $\vec{AB}$ and passing through some point, we need more information. If the plane should pass through the origin, then the plane will have equation:

$-2x+y+2z = 0$

3. Final Answer

a. Points $A$, $B$, and $C$ are plotted in 3D space.

b. $\vec{BA} = (2, -1, -2)$, $\vec{BC} = (0, 2, -1)$, $\vec{AC} = (-2, 3, 1)$

c. $\vec{BA} \cdot \vec{BC} = 0$, triangle $ABC$ is a right triangle.

d. The equation of the sphere is $x^2 + y^2 + z^2 - 4x - 2z - 4 = 0$.

e. Plane normal to $\vec{AB}$ through origin has equation: $-2x + y + 2z = 0$. Additional information is needed to determine a specific plane based on $\vec{AB}$.