The problem is to find the coordinates of vectors $CA$ and $CB$, calculate the dot product $CA \cdot CB$, and then find the angle $\angle ACB$. Also, show that triangle $ABC$ is a right-angled triangle, and find the equation of the plane $ABC$.

1. Problem Description

The problem is to find the coordinates of vectors $CA$ and $CB$, calculate the dot product $CA \cdot CB$, and then find the angle $\angle ACB$. Also, show that triangle $ABC$ is a right-angled triangle, and find the equation of the plane $ABC$.

2. Solution Steps

First, we are given the coordinates of the points $A(0, 1, 1)$, $B(2, 0, 0)$, and $C(0, 2, 0)$.

a. Find the vectors $CA$ and $CB$.

$CA = A - C = (0, 1, 1) - (0, 2, 0) = (0, -1, 1)$

$CB = B - C = (2, 0, 0) - (0, 2, 0) = (2, -2, 0)$

Now, compute the dot product $CA \cdot CB$:

$CA \cdot CB = (0)(2) + (-1)(-2) + (1)(0) = 0 + 2 + 0 = 2$

Next, find the magnitudes of the vectors $CA$ and $CB$:

$|CA| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2}$

$|CB| = \sqrt{2^2 + (-2)^2 + 0^2} = \sqrt{4 + 4 + 0} = \sqrt{8} = 2\sqrt{2}$

The angle $\angle ACB$ can be found using the formula:

$cos(\angle ACB) = \frac{CA \cdot CB}{|CA||CB|}$

$cos(\angle ACB) = \frac{2}{(\sqrt{2})(2\sqrt{2})} = \frac{2}{2(2)} = \frac{2}{4} = \frac{1}{2}$

Therefore, $\angle ACB = arccos(\frac{1}{2}) = 60^{\circ}$ or $\frac{\pi}{3}$ radians.

b. To show that $ABC$ is a right-angled triangle, we need to check if the Pythagorean theorem holds for the sides $AB, BC, CA$. First, compute vector $AB$:

$AB = B - A = (2, 0, 0) - (0, 1, 1) = (2, -1, -1)$

Then find magnitudes squared:

$|AB|^2 = 2^2 + (-1)^2 + (-1)^2 = 4 + 1 + 1 = 6$

$|BC|^2 = |CB|^2 = 8$

$|AC|^2 = |CA|^2 = 2$

Check for Pythagorean theorem: $AC^2 + BC^2 = AB^2$, or $BC^2 + AB^2 = AC^2$, or $AC^2 + AB^2 = BC^2$.

$2 + 8 = 10 \neq 6$

$8 + 6 = 14 \neq 2$

$2 + 6 = 8$, so $AC^2 + AB^2 = BC^2$

Hence, triangle $ABC$ is a right-angled triangle at $A$, since $AC^2 + AB^2 = BC^2$. The angle $\angle BAC = 90^{\circ}$.

c. Find the equation of the plane $ABC$.

Since we have vector $CA = (0, -1, 1)$ and vector $AB = (2, -1, -1)$, the normal vector $n$ to the plane can be found by taking the cross product of $CA$ and $AB$:

$n = CA \times AB = \begin{vmatrix} i & j & k \\ 0 & -1 & 1 \\ 2 & -1 & -1 \end{vmatrix} = i((-1)(-1) - (1)(-1)) - j((0)(-1) - (1)(2)) + k((0)(-1) - (-1)(2)) = i(1 + 1) - j(0 - 2) + k(0 + 2) = (2, 2, 2)$

We can simplify the normal vector to $n = (1, 1, 1)$. Using point A $(0, 1, 1)$ to find the equation of the plane:

$1(x - 0) + 1(y - 1) + 1(z - 1) = 0$

$x + y - 1 + z - 1 = 0$

$x + y + z = 2$

3. Final Answer

a. $CA = (0, -1, 1)$, $CB = (2, -2, 0)$, $CA \cdot CB = 2$, $\angle ACB = 60^{\circ}$ or $\frac{\pi}{3}$ radians.

b. Triangle $ABC$ is a right-angled triangle at $A$.

c. The equation of the plane $ABC$ is $x + y + z = 2$.