Show that the points A (1, –2, –8), B (5, 0, –2) and C (11, 3, 7) are collinear,

Solution:

The given points are A (1, –2, –8), B (5, 0, –2), and C (11, 3, 7).

$\therefore \overrightarrow{\mathrm{AB}}=(5-1) \hat{i}+(0+2) \hat{j}+(-2+8) \hat{k}=4 \hat{i}+2 \hat{j}+6 \hat{k}$

$\overrightarrow{\mathrm{BC}}=(11-5) \hat{i}+(3-0) \hat{j}+(7+2) \hat{k}=6 \hat{i}+3 \hat{j}+9 \hat{k}$

$\overrightarrow{\mathrm{AC}}=(11-1) \hat{i}+(3+2) \hat{j}+(7+8) \hat{k}=10 \hat{i}+5 \hat{j}+15 \hat{k}$

$|\overrightarrow{\mathrm{AB}}|=\sqrt{4^{2}+2^{2}+6^{2}}=\sqrt{16+4+36}=\sqrt{56}=2 \sqrt{14}$

$|\overrightarrow{\mathrm{BC}}|=\sqrt{6^{2}+3^{2}+9^{2}}=\sqrt{36+9+81}=\sqrt{126}=3 \sqrt{14}$

$|\overrightarrow{\mathrm{AC}}|=\sqrt{10^{2}+5^{2}+15^{2}}=\sqrt{100+25+225}=\sqrt{350}=5 \sqrt{14}$

$\therefore|\overrightarrow{\mathrm{AC}}|=|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|$

Thus, the given points A, B, and C are collinear.

Now, let point B divide AC in the ratio. Then, we have:

$\overrightarrow{\mathrm{OB}}=\frac{\lambda \overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OA}}}{(\lambda+1)}$

$\Rightarrow 5 \hat{i}-2 \hat{k}=\frac{\lambda(11 \hat{i}+3 \hat{j}+7 \hat{k})+(\hat{i}-2 \hat{j}-8 \hat{k})}{\lambda+1}$

$\Rightarrow(\lambda+1)(5 \hat{i}-2 \hat{k})=11 \lambda \hat{i}+3 \lambda \hat{j}+7 \lambda \hat{k}+\hat{i}-2 \hat{j}-8 \hat{k}$

$\Rightarrow 5(\lambda+1) \hat{i}-2(\lambda+1) \hat{k}=(11 \lambda+1) \hat{i}+(3 \lambda-2) \hat{j}+(7 \lambda-8) \hat{k}$

On equating the corresponding components, we get:

$5(\lambda+1)=11 \lambda+1$

$\Rightarrow 5 \lambda+5=11 \lambda+1$

$\Rightarrow 6 \lambda=4$

$\Rightarrow \lambda=\frac{4}{6}=\frac{2}{3}$

Hence, point B divides AC in the ratio