Show that the lines
  \[{{\overrightarrow{r}}_{1}}=3i+2j-4k+\lambda (i+2j+2k),\overrightarrow{{{r}_{2}}}=5i-2j+\mu (3i+2j+6k)\] are intersecting. Hence find their point of intersection.

Hint: We will first write both lines in components of i,j,k unit vectors as
 \[{{\overrightarrow{r}}_{1}}=(3+\lambda )i+(2+2\lambda )j+(2\lambda -4)k\] similarly, \[{{\overrightarrow{r}}_{2}}=(5+3\mu )i+(2\mu -2)j+6\mu k\]
Now if they are intersecting it means points are same so we will equate all three components of both lines respectively
$\left( 3+\lambda \right)i=\left( 5+3\mu \right)i$ , \[(2+2\lambda )j=(2\mu -2)j\] and \[(2\lambda -4)k=6\mu k\]
On comparing i component we get \[\lambda =2+3\mu \] and j component we get \[2\lambda =2\mu -4\]
On putting value we get \[\lambda =-4\] and \[\mu =-2\] , now putting these value back in any of the given lines we get the point of intersection as \[-i-6j-12k\]

Complete step-by-step answer:
Given two lines as
  \[{{\overrightarrow{r}}_{1}}=3i+2j-4k+\lambda (i+2j+2k),\overrightarrow{{{r}_{2}}}=5i-2j+\mu (3i+2j+6k)\] which are intersecting so we have to find their point of intersection. Here values of \[\lambda \] and \[\mu \] are unknown.
If the lines are intersecting it means we can equate all three components at a given value of \[\lambda \] and \[\mu \] from there we will calculate value of \[\lambda \] and \[\mu \]
Splitting in the components we can write \[{{\overrightarrow{r}}_{1}}=(3+\lambda )i+(2+2\lambda )j+(2\lambda -4)k\] and \[{{\overrightarrow{r}}_{2}}=(5+3\mu )i+(2\mu -2)j+6\mu k\] now as they are intersecting , equating each components respectively we get
 $\left( 3+\lambda \right)i=\left( 5+3\mu \right)i$, \[(2+2\lambda )j=(2\mu -2)j\] and \[(2\lambda -4)k=6\mu k\]
comparing i component we get \[\lambda =2+3\mu .......(1)\] and j component we get \[2\lambda =2\mu -4....(2)\]
putting equation (1) into the equation (2) we will get expression as \[2(2+3\mu )=2\mu -4\]
on expanding it looks like \[4+6\mu =2\mu -4\] ,on solving we get \[4+4\mu =-4\]
we get value of \[\mu \] equals to -2 , and putting it in equation (1) we get value of \[\lambda \] equals
 \[\lambda =2+3(-2)=2-6=-4\]
Now putting back values \[\lambda =-4\] and \[\mu =-2\] in any of the given line \[{{\overrightarrow{r}}_{1}}\] or \[\overrightarrow{{{r}_{2}}}\] we will get the intersection point.
 \[{{\overrightarrow{r}}_{2}}=(5+3\mu )i+(2\mu -2)j+6\mu k\] putting value of \[\mu \] in this equation we will get
 \[(5+3(-2))i+(2(-2)-2)j+6(-2)k\] which results into \[(-1)i+(-6)j+(-12)k\]
Hence intersection point is \[-i-6j-12k\]

Note: In this question we have equated only I and j component but we can consider any two component and equate it, we will get the same result for example we equate j and k component
 \[{{\overrightarrow{r}}_{1}}=(3+\lambda )i+(2+2\lambda )j+(2\lambda -4)k\]
, \[{{\overrightarrow{r}}_{2}}=(5+3\mu )i+(2\mu -2)j+6\mu k\]
Equating j component \[(2+2\lambda )=(2\mu -2)\] on dividing it by 2 gives \[\lambda =\mu -2\]
Equating K component \[(2\lambda -4)=6\mu \] on dividing it by 2 gives \[\lambda =2+3\mu \]
Equating both we again got \[\lambda =-4\] and \[\mu =-2\]