We know how to get the parameterization of a line segment so let’s just jump straight into the parameterization of the line segment.

\[\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {1,1,0} \right\rangle + t\left\langle {2,3, - 2} \right\rangle = \left\langle {1 + t,1 + 2t, - 2t} \right\rangle \,\,\,\,\,\,\,\,0 \le t \le 1\]

We could also break this up into parameter form as follows.

\[\begin{array}{*{20}{l}}{x = 1 + t}\\{y = 1 + 2t}\\{z = - 2t}\end{array}\hspace{0.25in}0 \le t \le 1\]

Either form of the parameterization will work for the problem but we’ll use the vector form for the rest of this problem.

We’ll need the magnitude of the derivative of the parameterization so let’s get that.

\[\vec r'\left( t \right) = \left\langle {1,2, - 2} \right\rangle \hspace{0.25in}\hspace{0.25in}\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2}} = \sqrt 9 = 3\]

We’ll also need the integrand “evaluated” at the parameterization. Recall all this means is we replace the \(x\)/\(y\) in the integrand with the \(x\)/\(y\) from parameterization. Here is the integrand evaluated at the parameterization.

\[xy - 4z = \left( {1 + t} \right)\left( {1 + 2t} \right) - 4\left( { - 2t} \right) = 2{t^2} + 11t + 1\]

The line integral is then,

\[\int\limits_{C}{{xy - 4z\,ds}} = \int_{0}^{1}{{\left( {2{t^2} + 11t + 1} \right)\left( 3 \right)\,dt}} = \left. {3\left( {\frac{2}{3}{t^3} + \frac{{11}}{2}{t^2} + t} \right)} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{43}}{2}}}\]