Parameterize C by the vector function
[tex]\left\langle x(t), y(t), z(t) \right\rangle = (1 - t) \left\langle 1, 0, 0 \right\rangle + t \left\langle 3, 1, 4\right\rangle[/tex]
[tex]\implies \left\langle x, y, z \right\rangle = \left\langle 1 + 2t, t, 4t \right\rangle[/tex]
with 0 ≤ t ≤ 1. Then computing differentials gives
[tex]\left\langle dx, dy, dz \right\rangle = \left\langle 2, 1, 4 \right\rangle \, dt[/tex]
So, the line integral is
[tex]\displaystyle \int_C z^2 \, dx + x^2 \, dy + y^2 \, dz = \int_C \left\langle z^2, x^2, y^2 \right\rangle \cdot \left\langle dx, dy, dz \right\rangle[/tex]
[tex]\displaystyle \int_C z^2 \, dx + x^2 \, dy + y^2 \, dz = \int_0^1 \left\langle (4t)^2, (1+2t)^2, t^2 \right\rangle \cdot \left\langle 2, 1, 4 \right\rangle[/tex]
[tex]\displaystyle \int_C z^2 \, dx + x^2 \, dy + y^2 \, dz = \int_0^1 (32t^2 + (1 + 2t)^2 + 4t^2) \, dt[/tex]
[tex]\displaystyle \int_C z^2 \, dx + x^2 \, dy + y^2 \, dz = \int_0^1 (40t^2 + 4t + 1) \, dt[/tex]
[tex]\displaystyle \int_C z^2 \, dx + x^2 \, dy + y^2 \, dz = \left(\frac{40}3\cdot1^3 + 2\cdot1^2 + 1\right) - \left(\frac{40}3\cdot0^3 + 2\cdot0^2 + 0\right)[/tex]
[tex]\displaystyle \int_C z^2 \, dx + x^2 \, dy + y^2 \, dz = \boxed{\frac{49}3}[/tex]
