First, we need to form two vectors on the plane to get the normal vector to the plane. This is done from the following operation:

\(\displaystyle v_1=p_1-p_2=\left \langle (2-3),(-5+1),(6-1) \right \rangle=\left \langle -1,-4,5\right \rangle\)

\(\displaystyle v_2=p_1-p_3=\left \langle (2-1),(-5+4),(6-1) \right \rangle=\left \langle 1,-1,5\right \rangle\)

We then take the cross product of these vectors, which gets us the normal vector

\(\displaystyle n=\left \langle -15,10,5\right \rangle\)

Plugging the point \(\displaystyle p_1\) and the normal vector into the equation of the plane, we get

\(\displaystyle -15(x-2)+10(y+5)+5(z-6)=0\)

Simplifying, we get

\(\displaystyle -15x+10y+5z=-50\)

Find the equation of the plane tangent to the surface \(\displaystyle z=f(x,y)\) at the point where \(\displaystyle x=x_o=\frac{\pi}{2}\) and \(\displaystyle y = y_o=2\). 

\(\displaystyle f(x,y)=\sin(2xy)\)                                                                                 (1) 

The equation of the plane tangent to the surface defined by \(\displaystyle f(x,y)\) is given by the formula: 

\(\displaystyle z=f(x_o,y_o)+f_x(x_o,y_o)(x-x_o)+f_y(x_o,y_o)(y-y_o)\)              (2)

In Equation (2) \(\displaystyle f_x\) and \(\displaystyle f_y\) are the partial derivatives of \(\displaystyle f\) with respect to \(\displaystyle x\) and \(\displaystyle y\), respectively. For this particular problem we have \(\displaystyle (x_o,y_o) =\left(\frac{\pi}{2},2\right)\) 

Let's fill in Equation (2) term-by-term: 

 \(\displaystyle f(x_o,y_o)=f\left(\frac{\pi}{2},2\right)=\sin(2\pi)=0\)

Compute the partial derivative and then evaluate both at \(\displaystyle \left(\frac{\pi}{2}, 2 \right )\) 

- Partial with respect to x 

\(\displaystyle f_x(x,y)=\frac{\partial f}{\partial x}=2y\cos(2xy)\)

\(\displaystyle f_x\left(\frac{\pi}{2}, 2 \right )=2(2)\cos(2\pi)=4\)

- Partial with respect to \(\displaystyle y\)

\(\displaystyle f_y(x,y)=2x\cos(2xy)\)

\(\displaystyle f_y\left(\frac{\pi}{2},2 \right )=2(2)\cos(2\pi) = 4\) 

Now fill in Equation (2) and simplify to get the equation of the tangent plane: 

\(\displaystyle z=4+4\left(x-\frac{\pi}{2}\right)+4(y-2)\)

\(\displaystyle =4x-2\pi+4y-8\)

\(\displaystyle =4x+4y-8-2\pi\)

Therefore the equation of the tangent plane to the surface \(\displaystyle z=\cos(2xy)\) at the point \(\displaystyle \left(\frac{\pi}{2},2 \right )\) is simply 

 \(\displaystyle z=4x+4y-8-2\pi\)

To find the equation of the plane, we use the formula \(\displaystyle A(x-x_0)+B(y-y_0)+C(z-z_0)=0\), where the point given is \(\displaystyle (x_0,y_0,z_0)\) and the normal vector \(\displaystyle n=\left \langle A,B,C\right \rangle\). Plugging in what we were given in the problem statement, we get \(\displaystyle 2(x+5)+1(y+4)-3(z-3)=0\). Manipulating the equation through algebra to isolate the variables, we get  \(\displaystyle 2x+y-3z=-23\).

There are several ways to solve this problem, but the most effective would be to notice that we can derive the following-

\(\displaystyle x^2 = 4\sin^2t\)

\(\displaystyle y^2 = 4\cos^2t\)

Hence

\(\displaystyle x^2+y^2 = 4\sin^2t + 4\cos^2t = 4\)

Therefore our curve is a circle of radius \(\displaystyle \sqrt4=2\), and it's circumfrence is \(\displaystyle 4\pi\). But we are only interested in half that circumfrence (\(\displaystyle t\) is from \(\displaystyle 0\) to \(\displaystyle \pi\), not \(\displaystyle 2\pi\).), so our answer is \(\displaystyle 2\pi\).

Alternatively, we could've found the length using the formula

\(\displaystyle l = \int_{t_0}^{t_1}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt\).

To find the coordinates, we set \(\displaystyle t=5\) into the curve function.

We get

\(\displaystyle c(5)=< 5^2,5,-3(5)>\)

and thus

\(\displaystyle c(5)=< 25,5,-15>\)

To find the coordinates, we evaluate the curve function for \(\displaystyle t=\pi/2\)

As such,

\(\displaystyle c(\frac{\pi}{2})=< cos (\frac{\pi}{2}), sin (\frac{\pi}{2}), cos (\frac{\pi}{2})>=< 0,1,0>\)

To find the equation of the line passing through these two points, we must first find the vector between them:

\(\displaystyle \vec{v}=\left \langle 1, 6, 2\right \rangle\)

This was done by finding the difference between the x, y, and z components for the vectors. (This can be done in either order, it doesn't matter.)

Now, pick a point to be used in the equation of the line, as the initial point. We write the equation of line as follows:

\(\displaystyle \vec{r}=\left \langle 1, 2, 3\right \rangle + t\left \langle 1, 6, 2\right \rangle=\left \langle 1+t, 2+6t, 3+2t\right \rangle\)

The choice of initial point is arbitrary. 

To find the coordinates of the parametric curve we plug in for

\(\displaystyle t=\pi/2\).

\(\displaystyle x=sin\,(\pi/2)= 1\)

\(\displaystyle y=(\pi/2)^2=\pi^2/4\)

As such the coordinates are

\(\displaystyle (1,\pi^2/4)\)

The velocity is given as the derivative of the position function, or

 \(\displaystyle v(t)=\frac{d(s(t))}{dt}\) .

We can use the quotient rule to find the derivative of the position function and then evaluate that at \(\displaystyle t=2\).  The quotient rule states that

 \(\displaystyle \left[\frac{f(x)}{g(x)}\right]'=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}\).  

In this case, \(\displaystyle f(t)=2t^2+1\Rightarrow f'(t)=4t\) and \(\displaystyle g(t)=t+1\Rightarrow g'(t)=1\).  

We can now substitute these values in to get

 \(\displaystyle \left[\frac{f(t)}{g(t)}\right]'=\frac{(t+1)*(4t)-(2t^2+1)*(1)}{(t+1)^2}=\frac{2t^2+4t-1}{(t+1)^2}\).  

Evalusting this at \(\displaystyle t=2\) gives us \(\displaystyle \frac{2*2^2+4*2-1}{(2+1)^2}=\frac{16-1}{3^2}=\frac{15}{9}\).  

So the answer is \(\displaystyle \frac{15}{9}\).