Using direction ratio shows that the points $A(2,3, - 4), B(1, - 2,3)$ and $C(3,8, - 11)$ are collinear.

Hint: First subtract 2 from 1, 3 from -2 and -4 from 3 to obtain the direction ratio $\left( {{a_1},{a_2},{a_3}} \right)$ of the line AB. Then subtract 1 from 3, -2 from 8 and 3 from -11 to obtain the direction ratio $\left( {{b_1},{b_2},{b_3}} \right)$ of the line BC. Then divide ${a_1}$ by ${b_1}$, ${a_2}$ by ${b_2}$ and ${a_3}$ by ${b_3}$, if all the ratios are same then the given points are collinear.

Formula Used:
If $\left( {{a_1},{a_2},{a_3}} \right)$are the direction cosines of a line AB and $\left( {{b_1},{b_2},{b_3}} \right)$ are the direction cosines of a line BC, then the points A, B and C are collinear if
$\dfrac{{{a_1}}}{{{b_1}}} = \dfrac{{{a_2}}}{{{b_2}}} = \dfrac{{{a_3}}}{{{b_3}}}$.

Complete step by step solution:
The direction ratios of the line formed by A, B are
$\left( {1 - 2, - 2 - 3,3 - ( - 4)} \right)$
=$\left( { - 1, - 5,7} \right)$
The direction ratios of the line formed by B, C are
$\left( {3 - 1,8 - ( - 2), - 11 - 3} \right)$
=$\left( {2,10, - 14} \right)$
Now, divide -1 by 2, -5 by 10 and 7 by -14 to observe whether the ratios are the same or not.
$ - \dfrac{1}{2}, - \dfrac{5}{{10}}, - \dfrac{7}{{14}}$
That is, $ - \dfrac{1}{2} = - \dfrac{1}{2} = - \dfrac{1}{2}$
As the ratios of the direction cosines of the line formed by A, B and B, C are the same,
Hence, A, B and C are collinear.

Note: We commonly obtain the equation by two points and then the direction cosines; this process is time-consuming but not incorrect. To answer the question quickly, obtain the direction cosines from the given points and their ratio.