Find a normal vector to the plane `x+2y+3z-6=0`

Explanation

Here, The given equation of the plane is x+2 y+3 z-6=0

`x+2 y+3 z=6`

`vec{r} cdot( vec{i}+2 vec{j}+3 vec{k})=6`

or, `{vec{r}}, {vec{n}}=6,`

where,` vec{n}= vec{i}+2 vec{j}+3 vec{k} ldots ldots `(i)

Now,

`|vec{n}|=sqrt{1^{2}+2^{2}+3^{2}}=sqrt{1+4+9}=sqrt{14}`

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