a) Find the symmetric and parametric equations of the line L1 through A(4, -2, 3)

parallel to the line L2: x = 2 - t, y = 3 + 5 t, z = -1 - 2t.

From L2, the coefficients of t yield the components of the vector a₂, which is parallel to the line L2.

This same vector can be used for L1 because the lines are parallel. Hence the parametric equations for L1 involve the given point and become x = 4 - t, y = -2 + 5t, z = 3 - 2t.

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 = 3 - 2t_xy to find the value of the parameter t for the x-y plane:

t_xy = 3/2.

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

x_xy = 4 - t_xy = 4 - 3/2 = 5/2.

y_xy = -2 + 5*t_xy = -2 + 5*(3/2) = 11/2

In other words, L1 intersects the x-y coordinate plane at (5/2, 11/2, 0).

For the x-z plane, we simply set y = 0 = -2 + 5t_xz to find the value of the parameter t for the x-z plane:

t_xz = 2/5.

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

x_xz = 4 - t_xz = 4 - 2/5 = 18/5.

z_xz = 3 - 2*t_xz = 3 - 2*(2/5) = 11/5

In other words, L1 intersects the x-z coordinate plane at (18/5, 0, 11/5).

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

t_yz = 4.

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

y_yz = -2 + 5t_yz = -2 + 5*4 = 18.

z_yz = 3 - 2*t_yz = 3 - 2*4 = -5

In other words, L1 intersects the x-z coordinate plane at (0, 18, -5).