Question:

Step 1: Determine the position vectors of $\vec{BA}$ and $\vec{BC}$. The position vector of A is $\vec{A} = (1, 2, 3)$. The position vector of B is $\vec{B} = (-1, 0, 0)$. The position vector of C is $\vec{C} = (0, 1, 2)$. The vector $\vec{BA}$ is given by $\vec{A} - \vec{B}$: $$ \vec{BA} = (1 - (-1), 2 - 0, 3 - 0) = (2, 2, 3) $$ The vector $\vec{BC}$ is given by $\vec{C} - \vec{B}$: $$ \vec{BC} = (0 - (-1), 1 - 0, 2 - 0) = (1, 1, 2) $$ Step 2: Calculate the dot product of $\vec{BA}$ and $\vec{BC}$. The dot product $\vec{BA} \cdot \vec{BC}$ is: $$ \vec{BA} \cdot \vec{BC} = (2)(1) + (2)(1) + (3)(2) = 2 + 2 + 6 = 10 $$ Step 3: Calculate the magnitudes of $\vec{BA}$ and $\vec{BC}$. The magnitude of $\vec{BA}$ is: $$ |\vec{BA}| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17} $$ The magnitude of $\vec{BC}$ is: $$ |\vec{BC}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} $$ Step 4: Use the dot product formula to find $\angle ABC$. The cosine of the angle $\theta = \angle ABC$ between two vectors $\vec{BA}$ and $\vec{BC}$ is given by: $$ \cos \theta = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| |\vec{BC}|} $$ Substitute the calculated values: $$ \cos \theta = \frac{10}{\sqrt{17} \cdot \sqrt{6}} = \frac{10}{\sqrt{17 \cdot 6}} = \frac{10}{\sqrt{102}} $$ Therefore, $\angle ABC$ is: $$ \angle ABC = \cos^{-1}\left(\frac{10}{\sqrt{102}}\right) $$