How to find parametric equations for the line through \[(2,4,6)\] that is perpendicular to the plane \[x-y+3z=7\] ?

Hint: We can solve the question using the concept of parametric equations.
Equation of plane: The plane containing the point \[\left( {{x}_{1}},{{y}_{_{1}}},{{z}_{1}} \right)\] and having a normal vector \[\overrightarrow{n}=\left\langle a,b,c \right\rangle \] has equation \[a\left( x-{{x}_{1}} \right)+b\left( y-{{y}_{1}} \right)+c\left( z-{{z}_{1}} \right)=0\]
V=< a,b,c > is the vector in the line. If the line is perpendicular to the plane, it is parallel to the normal vector of the plane, since the normal vector is also perpendicular to the plane.

Complete step-by-step answer:
According to our question the line \[x-y+z=7\] is perpendicular to the plane.
If a line is perpendicular to the plane then it is parallel to the Normal vector.
As our line is perpendicular to the plane, our line is also parallel to the Normal vector.
As our line is parallel to the Normal vector, the vector formed for our equation is \[\overrightarrow{n}=\left\langle 1,-1,3 \right\rangle \] .
Now our line parallel to the vector \[\overrightarrow{n}=\left\langle 1,-1,3 \right\rangle \] and passes through the point \[(2,4,6)\] .
A line parallel to the vector \[\overrightarrow{v}=\left\langle a,b,c \right\rangle \] passing through the point \[\left( {{x}_{1}},{{y}_{_{1}}},{{z}_{1}} \right)\] is given by the parametric equations.
\[\begin{align}
  & x={{x}_{1}}+at \\
 & y={{y}_{1}}+bt \\
 & z={{z}_{1}}+ct \\
\end{align}\]
From this we can see that our line is parallel to the vector \[\overrightarrow{n}=\left\langle 1,-1,3 \right\rangle \] and passing through the point \[(2,4,6)\] we can write the parametric equations as
\[\begin{align}
  & x=2+t \\
 & y=4-t \\
 & z=6+3t \\
\end{align}\]
From the above equations we can write the vector form as \[\overrightarrow{v}=\left\langle {{x}_{1}},{{y}_{1}},{{z}_{1}} \right\rangle +t\left\langle a,b,c \right\rangle \].
So from the above formula the vector form of our line formed \[\overrightarrow{r}=\left\langle 2,4,6 \right\rangle +t\left\langle 1,-1,3 \right\rangle \]
The vector form of the line is \[\overrightarrow{r}=\left\langle 2,4,6 \right\rangle +t\left\langle 1,-1,3 \right\rangle \] .

Note: We have to make sure we have done correctly while forming the vector and parametric equations, otherwise the vector form formed will be wrong. Knowledge on basic concepts of parametric equations will be more useful.