Using vectors,find the value of $k$ such that the points $(k,-10,3), (1,-1,3)$ and $(3,5,3)$ are collinear.

(D) Let the points be $A(k,-10,3), B(1,-1,3)$ and $C(3,5,3)$.
For points $A, B, C$ to be collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ must be parallel,i.e.,$\overrightarrow{AB} = \lambda \overrightarrow{BC}$ for some scalar $\lambda$.
$\overrightarrow{AB} = (1-k)\hat{i} + (-1 - (-10))\hat{j} + (3-3)\hat{k} = (1-k)\hat{i} + 9\hat{j} + 0\hat{k}$.
$\overrightarrow{BC} = (3-1)\hat{i} + (5 - (-1))\hat{j} + (3-3)\hat{k} = 2\hat{i} + 6\hat{j} + 0\hat{k}$.
Since $\overrightarrow{AB} = \lambda \overrightarrow{BC}$,we have:
$(1-k)\hat{i} + 9\hat{j} = \lambda(2\hat{i} + 6\hat{j})$.
Comparing the components:
$1-k = 2\lambda$ --- $(1)$
$9 = 6\lambda$ --- $(2)$
From $(2)$,$\lambda = \frac{9}{6} = \frac{3}{2}$.
Substituting $\lambda = \frac{3}{2}$ in $(1)$:
$1-k = 2 \times \frac{3}{2} = 3$.
$1-k = 3 \implies k = 1-3 = -2$.
Thus,the value of $k$ is $-2$.