Here's my pseudocode for all three parts of P24. This is not tested and may not be the best way, but I think the approach is solid.

x
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""" Return zero, one, or two triangles that result from clipping
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triangle abc with the x=1 plane."""
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function clip_positive_x(a::Vec3, b::Vec3, c::Vec3)
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  function in(v) = v[1] > 1 # helper - is v inside the x<1 half-space?
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  n_in = length([in(v) for v in (a, b, c)])
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  if n_in == 3
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    return a, b, c # case 1 - all in
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  elseif n_in == 0
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    return nothing # case 2 - all out
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  elseif n_in == 1
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    # case 3 - one in, two out; one triangle results
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    # yuck, need to handle different cases of which vertex is in; let's simplfiy:
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    # roll until the "in" vertex is a
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    while !in(a)
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        # if we were doing this for real, we might want to also keep track of
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        # how many times we rolled so we could return new vertices where the one
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        # inside is in the same position (i.e., "a" remains "a" if that one was
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        # inside); I'll gloss over this bookkeeping here
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        a, b, c = b, c, a # keep them in ccw order
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    end
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    bprime = int_x1(a, b)
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    cprime = int_x1(a, c)
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    return a, bprime, cprime
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  elseif n_in == 2
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    # similar procedure as above to get a and b to be the "in" vertices
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    while in(c)
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        a, b, c = b, c, a
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    end
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    ac = int_x1(a, c) 
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    bc = int_x1(b, c)
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    return [(a, b, ac), (b, bc, ac)]
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  end
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end
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​
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""" return where the line from p to q intersects the x = 1 plane """
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function int_x1(p, q)
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    qp = q-p
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    # phrase this as a ray: q + t (q-p)
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    # derive the t value:
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    #p.x + t * qp.x = 1
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    #t = (1 - p.x) / qp.x
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    t = (1 - p[1]) / qp[1]
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    return p + t * qp
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end