Find the equation of the plane passing through a point and a vector orthogonal

For a plane in $\mathbb{R}^3$ with $\mathbf{r_0}$ a point that lies in the plane and $\mathbf{n}$ a vector normal to the plane, its equation can be given by: $$ \mathbf{n}\cdot(\mathbf{r}-\mathbf{r_0})=0 \quad \mbox{where} \quad \mathbf{r}=(x,y,z) $$

To solve your question, let us first rearrange the above equation: $$ \mathbf{n}\cdot(\mathbf{r}-\mathbf{r_0})=0 \quad\Rightarrow\quad \mathbf{n}\cdot\mathbf{r}-\mathbf{n}\cdot\mathbf{r_0}=0 \quad\Rightarrow\quad \mathbf{n}\cdot\mathbf{r}=\mathbf{n}\cdot\mathbf{r_0} $$

Substituting values, gives the following equation for the plane in Cartesian form: $$ (1,7,-2)\cdot(x,y,z) = (1,7,-2)\cdot(3,1,6) \quad \Rightarrow \quad x+7y-2z = -2 $$

Key point:

A plane with normal vector $(n_1, n_2, n_3)$ has equation $$n_1 x + n_2 y + n_3 z = k$$ for some $k$.