Need help with finding equation of the plane containing the line and point. Given the symmetric equation.

Choose a point $A$ on the line first; If we set $z=0$, then 
$$\frac{5}{3}x-3 = \frac{4}{3}y + 1 = \frac{1}{3}(0) + 2=2.$$

Hence $x=3$, and $y=\frac{3}{4}$, and the point $A=(3,\frac{3}{4}, 0 )$ lies oin the line. Note that the vector
$$PA=A-P=(3,\frac{3}{4}, 0 )-(0,-1,-2)=(3,\frac{7}{4},2)$$ lies on the plane we are looking for.  Also the direction of the line is given by $v=(\frac{5}{3}, \frac{4}{3}, \frac{1}{3})$. The direction of the line can also be taken to be $w=3v=(5,4,1)$ which will make the computations easier. 

The normal vector of the plane should be orthogonal to both $PA$  and $w$. Thus the normal vector of the plane can be obtained by doing a cross-product, i.e.
\[n=PA\times v=(3,\frac{7}{4},2) \times (5,4,1)=\begin{vmatrix} i & j & k \\ 3 & \frac{7}{4} & 2 \\ 5 & 4 & 1 \end{vmatrix}  \]
\[=i(\frac{7}{4}-8)-j(3-10)+k(12-\frac{35}{4})=-\frac{25}{4}i+7j+\frac{13}{4}k.\]

So $n=-\frac{25}{4}i+7j+\frac{13}{4}k$. Equation of the plane with normal vector $n$ passing through the point $P (0,-1,-2)$ is 
\[-\frac{25}{4}(x)+7(y+1)+\frac{13}{4}(z+2)=0.\]