Answer
(a) $P(2,0,2)$
(b) $x+y=2$
Work Step by Step
(a) From the given points in the problem, we have
$x_1=1+t, y_1=1-t,z_1=2t$
and
$x_2=2-s, y_2=s,z_2=2$
Now, arranging $z's$ and $y's$, we have
$t=1$ and $s=0$
Putting these values in either parametric equation.
$x_1=1+1, y_1=1-1,z_1=2(1)$
Point of intersection: $P(2,0,2)$
(b) Equation of plane is defined as:
$a(x-x_0)+b(y-y_0)+c(z-z_0)$
Here, $\lt a,b,c\gt=\lt-2,-2,0\gt$ and $(x_0,y_0,z_0)=(2,0,2)$
$-2(x-2)+(-2)(y-0)+0(z-2)$
$-2x-4-2y=0$
$2x+2y=4$
Hence, $x+y=2$
