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Q. Statement I Points $(-4,6,10),(2,4,6)$ and $(14,0,-2)$ are collinear.
Statement II Point $(14,0,-2)$ divides the segment joining by other two given points in the ratio $3 : 2$ internally.

Introduction to Three Dimensional Geometry

Solution:

Let $A(-4,6,10), B(2,4,6)$ and $C(14,0,-2)$ be the given points. Let the point $P$ divides $A B$ in the ratio $k: 1$. Then, coordinates of the point $P$ are
$\frac{2 k-4}{k+1}, \frac{4 k+6}{k+1}, \frac{6 k+10}{k+1}$
Let us examine whether for some value of $k$, the point $P$ coincides with point $C$.
On putting $\frac{2 k-4}{k+1}=14$, we get $k=-\frac{3}{2}$
When $k=-\frac{3}{2}$, then $\frac{4 k+6}{k+1}=\frac{4\left(-\frac{3}{2}\right)+6}{-\frac{3}{2}+1}=0$
and $ \frac{6 k+10}{k+1}=\frac{6\left(-\frac{3}{2}\right)+10}{-\frac{3}{2}+1}=-2$
Therefore, $C(14,0,-2)$ is a point which divides $A B$ externally in the ratio $3: 2$ and is same as $P$. Hence, $A, B, C$ are collinear.