<s>[INST] Find the image of the point $(1,3,4)$ on the plane $2x-y+z+3=0$. 
Find the image of the point $(1,3,4)$ on the plane $2x-y+z+3=0$.

Let the image be $(a,b,c)$.
Equation of the line joining $(1,3,4)$ and  $(a,b,c)$ is $(a-1,b-3,c-4)$  and it will be parallel to the plane given.
Now we have $\dfrac{a-1}{2}=\dfrac{b-3}{-1}=\dfrac{c-4}{1}=k\implies a=2k+1,b=-k+3,c=k+4$.
Since  $(a,b,c)$ lies on the plane so we get $k=-1$.
Hence the point will be $(-1,4,3)$.
But the answer is not coming .Where am I wrong?? 
 [/INST] Let $P=a(1,3,4)$ then
$$2a-3a+4a+3=0\implies a=-1$$
thus $$P=(-1,-3,-4)$$
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