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

(N/A) The direction ratios of the line segment joining $A(2, 3, -4)$ and $B(1, -2, 3)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.
$a_1 = 1 - 2 = -1$
$b_1 = -2 - 3 = -5$
$c_1 = 3 - (-4) = 7$
So,the direction ratios of $AB$ are $(-1, -5, 7)$.
The direction ratios of the line segment joining $B(1, -2, 3)$ and $C(3, 8, -11)$ are:
$a_2 = 3 - 1 = 2$
$b_2 = 8 - (-2) = 10$
$c_2 = -11 - 3 = -14$
So,the direction ratios of $BC$ are $(2, 10, -14)$.
We observe that the direction ratios of $BC$ are $-2$ times the direction ratios of $AB$:
$(2, 10, -14) = -2 \times (-1, -5, 7)$.
Since the direction ratios are proportional,the lines $AB$ and $BC$ are parallel.
Because the point $B$ is common to both segments $AB$ and $BC$,the points $A, B,$ and $C$ must lie on the same straight line.
Therefore,the points $A, B,$ and $C$ are collinear.