How do I plot parametric equations in Matlab?

I have the attached equations for lines r & s. I have used the below code to plot r but I can not plot s.

I need to find the value of a knowing that both lines intersect at one point. Find the coordinates of the intersection point.

1. How do I plot 's'?

2. Is my working for plotting 'r' correct?

>> t=-10:0.5:10;
>> x=3-t;
>> y=1+2*t;
>> z=-4+3*t;
>> figure(1)
hold on
plot3(x,y,z,'xr')
plot3(x,y,z)
hold off
grid on
xlabel('x axis')
ylabel('y axis')
zlabel('z axis')

[Walter Roberson]

Given any particular lambda, your 3 equations for r lock down x, y, and z to single values. You can plot that easily.

Given any particular a, your 2 equations for s only give enough information to find two of the variables with relation to the third. That defines a line for each a rather than defining a point.

[FortuitousMonkey]

Could you expand on that please? For the line 'r' I have made t=-10:0.5:10 and plotted. For line 's' I know that it intersects 'r'. From that, can I find the value of a?

[Walter Roberson]

syms x y z a lambda
eqn = [x==3-lambda,y==1-2*lambda,z==-4+3*lambda,x+2*y+3*z==a,3*x-2*y+z==-a];
sol = solve(eqn);
sol.x, sol.y, sol.z, sol.lambda, sol.a

... all of which gets you one point, but is not enough to allow you to plot s independently as a line.

Consider

x + 2 * y + 3 * z == a
3 * x + 2 * y + z == -a

then from the first one,

x = a - 2 * y - 3 * z

substitute that into the second one:

3 * (a - 2 * y - 3 * z) + 2 * y + z == -a

which is

3*a - 4*y - 8*z == -a

which is

4*a - 4 * y - 8*z == 0

or

a - y - 2 * z == 0

for any given a value, that is only enough information to deduce

y = a - 2 * z

which gives you a relationship between y and z, but does not constrain z. You would need to plot using all z from -inf to +inf, and each z value would have a corresponding x and y point -- and this is the case for each a value.

[FortuitousMonkey]

Comment removed there was an error in my code.