Show that the vector-valued function shown below describes the function of a particle moving in a circle of radius 1 centered at a point (5,5,3) and lying in the plane 3x+3y-6z = 12

Part - 1 :

Given vector-valued function is  

                                   $r\left(t\right) = \left(5i +5j +3k\right) + \cos t \left(\frac{1}{\sqrt{2}}i-\frac{1}{\sqrt{2}}j\right)+\sin t\left(\frac{1}{\sqrt{3}}i+\frac{1}{\sqrt{3}}j+\frac{1}{\sqrt{3}}k\right)$

                                          $=\left(5+\frac{\cos t}{\sqrt{2}}+\frac{\sin t}{\sqrt{3}}\right) i +\left(5-\frac{\cos t}{\sqrt{2}}+\frac{\sin t}{\sqrt{3}}\right) j +\left(3+\frac{\sin t}{\sqrt{3}}\right) k$

      From this the parametric Equations of the given Curve are:

                        $x\left(t\right) = 5+\frac{\cos t}{\sqrt{2}}+\frac{\sin t}{\sqrt{3}}$

                       $y\left(t\right) = 5-\frac{\cos t}{\sqrt{2}}+\frac{\sin t}{\sqrt{3}}$

                       $z\left(t\right) = 3+\frac{\sin t}{\sqrt{3}}$

   Substitute these in the plane equation 3x+3y-6z = 12

                    $3 \left( 5+\frac{\cos t}{\sqrt{2}}+\frac{\sin t}{\sqrt{3}}\right) + 3 \left( 5-\frac{\cos t}{\sqrt{2}}+\frac{\sin t}{\sqrt{3}}\right) - 6\left( 3+\frac{\sin t}{\sqrt{3}}\right) =12$

         $\Rightarrow$   $\left(15+15-18\right) + \left(\frac{3}{\sqrt{2}}-\frac{3}{\sqrt{2}}\right) \cos t +\left(\frac{3}{\sqrt{3}}+\frac{3}{\sqrt{3}}-\frac{6}{\sqrt{3}}\right) =12$

        $\Rightarrow$    $12 + \left(0\right) + \left(0\right) =12$

        $\Rightarrow$    12 = 12

  So the given vector-valued function is lying in the plane 3x+3y-6z = 12 .