Find the image of the line $\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ in the plane $x+y+z=4$. 
Find the image of the line $\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ with respect to plane $x+y+z=4$.

my method:
Any point in the given line will be $(3t,2t,t+1)$.
Thus the line meets the plane in $(3/2,1,3/2)$.
We now take a random point $(0,0,1)$ on this line. Clearly its image  will lie on lines image wrt to this plane.
We have to find image of $(0,0,1)$.
Since the perpendicular to the plane passing through $(0,0,1)$ is $$x=y=z-1$$. It will meet the plane in $(1,1,2)$. i.e image is $(2,2,3)$.
Thus the image line which passes through $(3/2,1,3/2)$ ad $(2,2,3)$ is $$\frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3}$$

I  am  however looking for alternative methods
 A: The line $r:\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ is the intersection of the planes $\alpha:\frac{x}{3}=\frac{y}{2};\;\beta:\frac{x}{3}=z-1$.
Any linear combination $\mathcal{P}$ of the two planes $\alpha:2x-3y=0$ and $\beta:x-3z+3=0$ passes through the line $r$
$$\mathcal{P}:h(2x-3y)+k(x-3z+3)=0;\;\forall h,k\in\mathbb{R}\land(h^2+k^2\ne 0)$$
Reorder the equation of $\mathcal{P}$
$$\mathcal{P}:x(2h+k)-3hy-3kz+3k=0$$
We have the plane passing through $r$ perpendicular to the given plane $\pi:x+y+z=4$ if the normal vectors $p=(2h+k,-3h,-3k)$ and $u=(1,1,1)$ are orthogonal, that is $p\cdot u=0$, which means
$$2h+k-3h-3k=0\to h=-2k$$
Plug in the equation of $\mathcal{P}$
$$\mathcal{P*}:x(-4k+k)+6ky-3kz+3k=0\to -3k(x-2y+z-1)=0$$
The plane perpendicular to $\pi$ passing through $r$ has equation $\gamma:x-2y+z=1$.
Thus the projection $r_{\pi}$ of $r$ on $\pi$ is the line
$$\begin{cases}
x+y+z=4\\
x-2y+z=1\\
\end{cases}
$$
that is $$r_{\pi}:\left( 3-t,1,t\right)$$
or
$$
\begin{cases}
x=3-z\\
y=1\\
\end{cases}
$$
A: An alternate method. The plane has normal vector $n=\langle 1,1,1\rangle$. The line's direction vector is $v=\langle3,2,1\rangle$. So decompose the direction vector $v$ into its two components, one parallel to $n$ and one orthogonal.
$$v_{\text{parallel to $n$}}=\frac{v\cdot n}{n\cdot n}n=\frac{6}{3}n=\langle 2,2,2\rangle$$
$$v_{\text{orthogonal to $n$}}=v-v_{\text{parallel}}=\langle 1,0,-1\rangle$$
Geometrically, the direction vector of the image line should be $v_{\text{orthogonal to $n$}}-v_{\text{parallel to $n$}}=\langle-1,-2,-3\rangle$. And as you found, the image line also passes through $(3/2,1,3/2)$. So a parametric formula for the line is $(3/2-t,1-2t,3/2-3t)$. Or in symmetric form (solving for $t$ in the three coordinates): $$\frac32-x=\frac{1-y}{2}=\frac{3/2-z}{3}$$
