a) Find the symmetric and parametric equations of the line L1 through A(-1, 3, 2)

perpendicular to the plane 3x - 2y + 4z = 5.

From the equation of the plane, the coefficients yield the components of the normal vector (which is perpendicular to the plane and hence can be used in the equation of the desired line).

n = 3i - 2j + 4k.

Hence the parametric equations for L1 involve the given point and this vector, becoming

x = -1 + 3t, y = 3 - 2t, z = 2 + 4t.

The symmetric equations for L1 can be found by solving this last equation for t:

b) Find the intersection of the line L1 with the coordinate planes.

For the x-y plane, we simply set z = 0 = 2 + 4t_xy to find the value of the parameter t for the x-y plane:

t_xy = -1/2.

Now we use this value of t in the equations for x and y:

x_xy = -1 + 3t_xy = -1 + 3(-1/2) = -5/2.

y_xy = 3 - 2*t_xy = 3 -2(-1/2) = 4

In other words, L1 intersects the x-y coordinate plane at (5-/2, 4, 0).

For the x-z plane, we simply set y = 0 = 3 - 2t_xz to find the value of the parameter t for the x-z plane:

t_xz = 3/2.

Now we use this value of t in the equations for x and z:

x_xz = -1 + 3t_xz = -1 + 3(3/2) = 7/2.

z_xz = 2 + 4*t_xz = 2 + 4*(3/2) = 8

In other words, L1 intersects the x-z coordinate plane at (7/2, 0, 8).

For the y-z plane, we simply set x = 0 = -1 + 3t_yz to find the value of the parameter t for the y-z plane:

t_yz = 1/3.

Now we use this value of t in the equations for y and z:

y_yz = 3 - 2t_yz = 3 - 2*(1/3) = 7/3.

z_yz = 2 + 4*t_yz = 2 + 4*(1/3) = 10/3

In other words, L1 intersects the x-z coordinate plane at (0, 7/3, 10/3).