If $$O,\,P$$  are the points $$\left( {0,\,0,\,0} \right),\,\left( {2,\,3,\, - 1} \right)$$     respectively, then what is the equation to the plane through $$P$$ at right angles to $$OP\,?$$

Solution :
Since, coordinates of points $$O$$ and $$P$$ are $$\left( {0,\,0,\,0} \right)$$   and $$\left( {2,\,3,\, - 1} \right)$$   respectively.
Direction ratios of $$OP$$  are $$\left\langle {2,\,3,\, - 1} \right\rangle $$
The plane is perpendicular to $$OP.$$
So, its equation is $$2x + 3y - z + d = 0......\left( {\text{i}} \right)$$
Since, this plane passes through $$\left( {2,\,3,\, - 1} \right);$$
$$\eqalign{ & 2 \times 2 + 3 \times 3 - 1 \times - 1 + d = 0 \cr & \Rightarrow 4 + 9 + 1 + d = 0 \cr & \Rightarrow d = - 14 \cr} $$
On putting the value of $$d$$ in equation $$\left( {\text{i}} \right)$$
$$\eqalign{ & 2x + 3y - z - 14 = 0 \cr & \Rightarrow 2x + 3y - z = 14 \cr} $$
which is required equation of plane.