Find the angles between the lines.
\[\begin{array}{l}{{\vec r}_1} = 2\hat i + 3\hat j - 4\hat k + t\left( {\hat i - 2\hat j + 2\hat k} \right)\\{{\vec r}_2} = 3\hat i - 5\hat k + t\left( {3\hat i - 2\hat j + 6\hat k} \right)\end{array}\]

Hint:
Here we will first compare the given equation of lines with the standard vector equation of lines to get their parallel vectors. Then we will find the dot product between these parallel vectors of the given lines and solve it to get the value of the angle between them.

Complete step by step solution:
The given equations of the lines are \[{\vec r_1} = 2\hat i + 3\hat j - 4\hat k + t\left( {\hat i - 2\hat j + 2\hat k} \right),{\vec r_2} = 3\hat i - 5\hat k + t\left( {3\hat i - 2\hat j + 6\hat k} \right)\].
First, we will compare this equation of the lines to the standard equation of lines i.e. \[\vec u = \vec a + \lambda \vec b\] where, \[\vec a\] is the vector of the point through which the line is passing and \[\vec b\] is the parallel vector of the line. Therefore, from this we will get the value of the parallel vectors of the given lines i.e. \[{\vec b_1}\& {\vec b_2}\].
\[\begin{array}{l}{{\vec b}_1} = \hat i - 2\hat j + 2\hat k\\{{\vec b}_2} = 3\hat i - 2\hat j + 6\hat k\end{array}\]
Let the angle between the lines be \[\theta \]. Therefore, we will find the dot product of the parallel vectors of the lines and solve it to get the value of \[\theta \]. Therefore, we get
\[{\vec b_1} \cdot {\vec b_2} = \left| {{{\vec b}_1}} \right| \cdot \left| {{{\vec b}_2}} \right|\cos \theta \]
Now putting the values in the above equation, we get
\[ \Rightarrow \left( {\hat i - 2\hat j + 2\hat k} \right) \cdot \left( {3\hat i - 2\hat j + 6\hat k} \right) = \left( {\sqrt {1 + 4 + 4} } \right) \cdot \left( {\sqrt {9 + 4 + 36} } \right)\cos \theta \]
Multiplying the vectors, we get
\[ \Rightarrow 3 + 4 + 12 = \left( {\sqrt {1 + 4 + 4} } \right) \cdot \left( {\sqrt {9 + 4 + 36} } \right)\cos \theta \]
Now we will solve this further and we will keep the term with \[\theta \] on one side of the equation.
Adding the terms in the bracket, we get
\[ \Rightarrow 19 = \left( {\sqrt 9 } \right) \cdot \left( {\sqrt {49} } \right)\cos \theta \]
Simplifying the equation, we get
\[ \Rightarrow 19 = \left( 3 \right) \cdot \left( 7 \right)\cos \theta \]
Multiplying the terms, we get
\[ \Rightarrow 19 = 21 \times \cos \theta \]
Dividing both side by 21, we get
\[ \Rightarrow \cos \theta = \dfrac{{19}}{{21}}\]
Now taking the inverse of the cos function to get the value of \[\theta \]. Therefore, we get
\[ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{19}}{{21}}} \right)\]

Hence, the angle between the lines, \[\theta = {\cos ^{ - 1}}\left( {\dfrac{{19}}{{21}}} \right)\].

Note:
Vector is the geometric object that has both the magnitude and the direction of an object. So while calculating the equation of a line vector we should know that it is equal to the difference between the final point vector and the starting point vector of that line. We also know that Vectors have three components i.e. \[x\] component, \[y\] component and \[z\] component and all the three components of the vectors are perpendicular to each other. Unit vector is a vector which has a magnitude of 1 unit and zero vector is a vector which has a magnitude of 0 unit.