How do you find the parametric and symmetric equation of the line passing through the point (2, 3, 4) and perpendicular to the plane 5x+6y-7z=20?

Hint: For solving this equation first of all we should be aware of parametric and symmetric equations. In the vector form of the line we get a position vector for the point and in
the parametric form we get the actual coordinates of the point. This is called the symmetric equations of the line

Complete Step by Step answer:
For finding a parametric equation
a. First of all we have to find a set of the parametric equations for the given equation y=x+3.
b. Now, we will assign any of a variable equal to t.
c. After that we will write the given equation as y=t+3.
d. Therefore, a set of parametric equations is x = t and y=t+3.

By using the fact that a vector is always perpendicular to a plane Ax+By+Cz=D 

is: $A\hat i\, + \,B\hat j\, + \,C\hat k$

Point-Vector which is normal to the plane, 5x+6y−7z=20, is:

$5\hat i\, + \,6\hat j\, + \,7\hat k$
The vector form of the line which passes through the point (2, 3, 4):

(x ,y ,z) = (2, 3, 4) + t$\left( {5\hat i\, + \,6\hat j\, + \,7\hat k} \right)$

By using the point-vector form we have find the 3 parametric equations which are:
x=5t+2 y=6t+3 z=−7t+4

Now, the symmetric form of the equation by solving parametric form are:

$t\, = \,\dfrac{{x - 2}}{3}$ , $t\, = \,\dfrac{{y - 3}}{6}$ and $t\, = \,\dfrac{{z - 4}}{{ - 7}}$

The symmetric form of equation is


$\dfrac{{x - 2}}{3}\, = \,\,\dfrac{{y - 3}}{6}\,\, = \,\,\dfrac{{z - 4}}{{ - 7}}$

Note: The parametric equations allow you to make a graph by providing us the details such as height, direction, and speed. The symmetric form of the equation represents the two variables x and y in relationship to the x-intercept and the y-intercept b.