The problem asks to show that the points $A(-2i + 3j + 5k)$, $B(i + 2j + 3k)$, and $C(7i - k)$ are collinear.

1. Problem Description

The problem asks to show that the points $A(-2i + 3j + 5k)$, $B(i + 2j + 3k)$, and $C(7i - k)$ are collinear.

2. Solution Steps

To show that three points A, B, and C are collinear, we need to show that the vectors $\vec{AB}$ and $\vec{AC}$ are parallel. This means that one vector is a scalar multiple of the other.

First, let's find the vectors $\vec{AB}$ and $\vec{AC}$.

$\vec{AB} = B - A = (i + 2j + 3k) - (-2i + 3j + 5k) = (1 - (-2))i + (2 - 3)j + (3 - 5)k = 3i - j - 2k$

$\vec{AC} = C - A = (7i - k) - (-2i + 3j + 5k) = (7 - (-2))i + (0 - 3)j + (-1 - 5)k = 9i - 3j - 6k$

Now, we check if $\vec{AC}$ is a scalar multiple of $\vec{AB}$.

$\vec{AC} = k \vec{AB}$

$9i - 3j - 6k = k(3i - j - 2k)$

$9i - 3j - 6k = 3ki - kj - 2kk$

Comparing the coefficients of $i$, $j$, and $k$, we have:

$9 = 3k \Rightarrow k = 3$

$-3 = -k \Rightarrow k = 3$

$-6 = -2k \Rightarrow k = 3$

Since the value of $k$ is the same for all components ($k = 3$), $\vec{AC} = 3\vec{AB}$.

Therefore, $\vec{AB}$ and $\vec{AC}$ are parallel, and since they share the common point A, the points A, B, and C are collinear.

3. Final Answer

The points A, B, and C are collinear.