Equation of plane containing the line $\frac{x-1}{-3}=\frac{y-3}{2}=\frac{z-2}{1}$ and the point $\left(0,-3,4\right)$ is

Equation of the given line is $\frac{x-1}{-3}=\frac{y-3}{2}=\frac{z-2}{1}$ and the given point is $\left(0,-3,4\right)$

Points on the plane are $A\left(1,3,2\right)$ and $B\left(0,-3,4\right)$

Vectors along the plane are $\stackrel{\rightharpoonup }{AB}=i+6j-2k$ and $-3i+2j+k$

Cross product of these two vectors is normal vector to the required plane

The normal vector to the plane is $\begin{array}{rcl}\left|\begin{array}{ccc}i& j& k\\ 1& 6& -2\\ -3& 2& 1\end{array}\right|& =& i\left(10\right)-j\left(-5\right)+k\left(20\right)\\ & =& 5\left(2i+j+4k\right)\end{array}$

Hence, the equaiton of required plane is

 $\begin{array}{rcl}2\left(x-1\right)+\left(y-3\right)+4\left(z-2\right)& =& 0\\ 2x-2+y-3+4z-8& =& 0\\ 2x+y+4z-13& =& 0\end{array}$