Equation of right circular cylinder with radius of the base as 2 units. Obtain the equation of right circular cylinder with radius of the base as 2 units. Its axis passes through $(1, 2, 3)$ and direction cosines are given as $(2, -3, 6)$
I got $45x^2+40y^2+13z^2+12xy-36yz-24zx-42x-280y-126z+294 = 0$

Put
$$
\begin{gathered}
  A = \left( {1,2,3} \right) \hfill \\
  X = \left( {x,y,z} \right) \hfill \\
  \mathbf{n} = \frac{1}
{7}\left( {2, - 3,6} \right) \hfill \\ 
\end{gathered} 
$$
Then the required cylinder is the lieu of points such that:
$$
2 = \left| {\mathop {AX}\limits^ \to   \times \mathbf{n}} \right|
$$
that is:
$$
14 = \left| {\left( {6\left( {y - 2} \right) + 3\left( {z - 3} \right),\; - 6\left( {x - 1} \right) + 2\left( {z - 3} \right),\; - 3\left( {x - 1} \right) - 2\left( {y - 2} \right)} \right)} \right|
$$
i.e.
$$
\left( {6\left( {y - 2} \right) + 3\left( {z - 3} \right)} \right)^2  + \left( { - 6\left( {x - 1} \right) + 2\left( {z - 3} \right)} \right)^2  + \left( { - 3\left( {x - 1} \right) - 2\left( {y - 2} \right)} \right)^2  - 14^2  = 0
$$
which confirm your equation, except for the sign of 
$36yz$ which shall be positive, not negative.
