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

Here, use the following stepwise approach first, get the values of

$|\overrightarrow {{\rm{AB}}} |,\mid \left| {\overrightarrow {{\rm{BC}}} } \right|$ and $|\overrightarrow {{\rm{AC}}} |$

if $|\overrightarrow {{\rm{AB}}} | + |\overrightarrow {{\rm{BC}}} | = |\overrightarrow {{\rm{AC}}} |$ such that.

Let the points are $A(k, - 10,3),B(1, - 1,3)$ and $C(3,5,3)$.

So, $\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} - \overrightarrow {{\rm{OA}}}$

$= (\widehat {\rm{i}} - \widehat {\rm{j}} + 3\widehat {\rm{k}}) - (k\widehat {\rm{i}} - 10\widehat {\rm{j}} + 3\widehat {\rm{k}})$

$= (1 - k)\widehat {\rm{i}} + ( - 1 + 10)\widehat {\rm{j}} + (3 - 3)\widehat {\rm{k}}$

$= (1 - k)\widehat {\rm{i}} + 9\widehat {\rm{j}} + 0\widehat {\rm{k}}$

$\therefore$ $|\overrightarrow {{\rm{AB}}} | = \sqrt {{{(1 - k)}^2} + {{(9)}^2} + 0} = \sqrt {{{(1 - k)}^2} + 81}$

Similarly, $\overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{OC}}} - \overrightarrow {{\rm{OB}}}$

$= (3\widehat {\rm{i}} + 5\widehat {\rm{j}} + 3\widehat {\rm{k}}) - (\widehat {\rm{i}} - \widehat {\rm{j}} + 3\widehat {\rm{k}})$

$= 2\widehat {\rm{i}} + 6\widehat {\rm{j}} + 0\widehat {\rm{k}}$
$\therefore$ $|\overrightarrow {{\rm{BC}}} | = \sqrt {{2^2} + {6^2} + 0} = 2\sqrt {10}$

and $\overrightarrow {{\rm{AC}}} = \overrightarrow {{\rm{OC}}} - \overrightarrow {{\rm{OA}}}$

$= (3\widehat {\rm{i}} + 5\widehat {\rm{j}} + 3\widehat {\rm{k}}) - (k\widehat {\rm{i}} - 10\widehat {\rm{j}} + 3\widehat {\rm{k}})$

$= (3 - k)\widehat {\rm{i}} + 15\widehat {\rm{j}} + 0\widehat {\rm{k}}$

$\therefore$ $|{\rm{A\vec C}}| = \sqrt {{{(3 - k)}^2} + 225}$

As per the property, if $A,B$ and $C$ are collinear then sum of modulus of any two vectors will be equal to the modulus of third vectors

For $|\overrightarrow {AB} | + |\overrightarrow {BC} | = |\overrightarrow {AC} |$

$\sqrt {{{(1 - k)}^2} + 81} + 2\sqrt {10} = \sqrt {{{(3 - k)}^2} + 225}$

$\Rightarrow$ $\sqrt {{{(3 - k)}^2} + 225} - \sqrt {{{(1 - k)}^2} + 81} = 2\sqrt {10}$

$\Rightarrow$ $\sqrt {9 + {k^2} - 6k + 225} - \sqrt {1 + {k^2} - 2k + 81} = 2\sqrt {10}$

$\Rightarrow$ $\sqrt {{k^2} - 6k + 234} - 2\sqrt {10} = \sqrt {{k^2} - 2k + 82}$

$\Rightarrow$ ${k^2} - 6k + 234 + 40 - 2\sqrt {{k^2} - 6k + 234} \cdot 2\sqrt {10} = {k^2} - 2k + 82$

$\Rightarrow$ ${k^2} - 6k + 234 + 40 - {k^2} + 2k - 82 = 4\sqrt {10} \sqrt {{k^2} + 234 - 6k}$

$\Rightarrow$ $- 4k + 192 = 4\sqrt {10} \sqrt {{k^2} + 234 - 6k}$

$\Rightarrow$ $- k + 48 = \sqrt {10} \sqrt {{k^2} + 234 - 6k}$

Squaring both sides we get

$48 \times 48 + {k^2} - 96k = 10\left( {{k^2} + 234 - 6k} \right)$
$\Rightarrow$ ${k^2} - 96k - 10{k^2} + 60k = - 48 \times 48 + 2340$

$\Rightarrow$ $- 9{k^2} - 36k = - 48 \times 48 + 2340$

$\Rightarrow$ $\left( {{k^2} + 4k} \right) = + 16 \times 16 - 260\quad$ [dividing by 9 in both sides]

$\Rightarrow$ ${k^2} + 4k = - 4$
${k^2} + 4k + 4 = 0$

$\Rightarrow$ ${(k + 2)^2} = 0$
$\therefore$ $k = - 2$