Projection from point onto plane

Question

Let the plane is $\prod:\vec{x}\bullet\hat{n}=d$ be a plane, where $\vec{x}\in R^{3}$ and $d\in R$, then can I show using norm and Cauchy-Schwarz inequality that projection of point $x$ onto the plane is equal $$Proj_{\prod}(\vec{x})=\vec{x}+(d-\vec{x}\bullet\hat{n})\hat{n}.$$ Let $\vec{z}$ be any point on the plane then $\vec{z}\bullet\hat{n}=d$, now the distance between $\vec{z}$ and any point $\vec{x}\in R^3$ is $$||\vec{z}-\vec{x}||$$ and $$\Rightarrow ~\big(||\vec{z}-\vec{x}||\big)^2=\big(||\hat{n}||\big)^2 \big(||\vec{z}-\vec{x}||\big)^2 =\big(||\hat{n}\times(\vec{z}-\vec{x})||\big)^2+|\hat{n}\bullet(\vec{z}-\vec{x})|^2$$,where ||\hat{n}||=1. I have problem in the above equality, I know that the Cauchy-Schwarz inequality is $$|<\vec{x},\vec{y}>|\leq||\vec{x}||||\vec{y}||$$. How can I use this Cauchy-Schwarz inequality in the above case that $$\big(||\hat{n}\times(\vec{z}-\vec{x})||\big)^2+|\hat{n}\bullet(\vec{z}-\vec{x})|^2\geq|\hat{n}\bullet(\vec{z}-\vec{x})|^2$$ The equality holds only $iff$ $\hat{n}\times(\vec{y}-\vec{x})=0$ I can proceed further to get the desired projection.