tough question in Line integral / Multivariable Calculas 
Let $C$ be a curve in the $(x − y)$-plane. For every point $(x, y)$ of $C$
  let $u(x, y)$ denote the unit vector in the direction of the tangent line
  to $C$ at $(x, y)$. Let $S$ be the surface obtained by taking the union of
  all straight line segments connecting $(1, 2, 3)$ to points of $C$. Express
  the area of $S$ as an integral of the first type, on the curve $C$, of some
  function of $x$ and $y$. (hint: try to use the function $u(x, y)$.)

really Hard question , i couldn't understand how to use the fact that line integral will help here since i don't have a function $f(x,y)$ to calculate $ \int_{C} f(x(t),y(t))\sqrt{x'(t)^2 + y'(t)^2 }\ dt$ also i can parameterize $S$ like that :
$S=:k(1-x(t),2-y(t),3) , k\in[0,1]$ where $(x(t),y(t),0)$ is the curve $C$
and why $u(x,y)$ is given here.
Unit vector 

$c(t) = (x(t),y(t), 0)\\
\frac {dc}{dt} = u(t)$
The parmeterization of $S$ should be
$S = (k +(1-k) x,2 k + (1-k) y, 3k)$
or
$((1-k) +k x,2(1- k) + k) y(t), 3(1-k))$
with 
$0\le k \le 1$
$dS = \|\frac {\partial S}{dk} \times \frac {\partial S}{dt}\|$
$\frac {\partial S}{dk} = (1-x, 2-y,3) = (1,2,3) - c(t)$
$\frac {\partial S}{dt} = (1-k)u(t)$
$\int_0^1 (1-k) dk \int_0^t \|(1,2,3)\times u(t) - c(t)\times u(t)\| \ dt$
$\frac 12 \int_0^t \|(1,2,3)\times u(t) - c(t)\times u(t)\| \ dt$
