$$P\left( {a,\,b,\,c} \right);\,Q\left( {a + 2,\,b + 2,\,c - 2} \right)$$       and $$R\left( {a + 6,\,b + 6,\,c - 6} \right)$$     are collinear.
Consider the following statements :
1. $$R$$ divides $$PQ$$  internally in the ratio $$3 : 2$$
2. $$R$$ divides $$PQ$$  externally in the ratio $$3 : 2$$
3. $$Q$$ divides $$PR$$  internally in the ratio $$1 : 2$$
Which of the statements given above is/are correct ?

Solution :
Given that $$P\left( {a,\,b,\,c} \right);\,Q\left( {a + 2,\,b + 2,\,c - 2} \right)$$       and $$R\left( {a + 6,\,b + 6,\,c - 6} \right)$$     are collinear, one point must divide, the other two points externally or internally.
Let $$R$$ divide $$P$$ and $$Q$$ in ratio $$k : 1$$
So, taking on $$x$$-co-ordinates
$$\eqalign{ & \frac{{k\left( {a + 2} \right) + a}}{{k + 1}} = a + 6 \cr & \Rightarrow k\left( {a + 2} \right) + a = \left( {k + 1} \right)\left( {a + 6} \right) \cr & \Rightarrow ka + 2k + a = ka + 6k + a + 6 \cr & \Rightarrow - 4k = 6{\text{ or }}k = - \frac{3}{2} \cr} $$
Negative sign shows that this is external division in ratio $$3 : 2.$$
So, $$R$$ is divided $$P$$ and $$Q$$ externally in $$3 : 2$$  ratio.
Putting this value for $$y$$ - and $$z$$ - coordinates satisfied :
for $$y$$ - co-rdinate :
$$\frac{{3\left( {b + 2} \right) - 2b}}{{3 - 2}} = \frac{{3b + 6 - 2b}}{1} = b + 6$$
and for $$z$$-co-rdinate :
$$\frac{{3\left( {c - 2} \right) - 2c}}{{3 - 2}} = \frac{{3c - 6 - 2c}}{1} = c - b$$
Statement $$\left( 2 \right)$$ is correct.
Also, let $$Q$$ divide $$P$$ and $$R$$ in ratio $$P : 1$$
Taking an $$x$$-co-ordinate :
$$\eqalign{ & \frac{{p\left( {a + 6} \right) + a}}{{p + 1}} = a + 2 \cr & \Rightarrow \frac{{p.a + 6p + a}}{{p + 1}} = a + 2 \cr & \Rightarrow pa + 6p + a = pa + a + 2p + 2 \cr & \Rightarrow 4p = 2 \cr & \Rightarrow p = \frac{1}{2} \cr} $$
Positive sign shows but the division is internal and in the ratio $$1 : 2$$
Verifying for $$y$$-and $$z$$-co-ordinates, satisfies this results.
For $$y$$- co-ordinate,
$$\frac{{\left( {b + 6} \right) \times 1 + 2b}}{3} = \frac{{3b + 6}}{3} = b + 2$$
and for $$z$$-co-ordinate,
$$\frac{{c - 6 + 2c}}{3} = \frac{{3c - 6}}{3} = c - 2$$
values are satisfied. So, statement $$\left( 3 \right)$$ is correct.