Find the distance $d$ from the point $( 2,3,-1)$ to the plane $x+4y+z=5$.

Solution We use (1) with $A=1$, $B=4$, $C=1$, $D=5$, and $P_{0}=(2,3,-1)$. Then $$d=\frac{\left\vert (1)(2)+(4)(3)+(1)(-1)-5\right\vert }{\sqrt{1+16+1}}= \frac{8}{3\sqrt{2}}=\frac{4\sqrt{2}}{3}$$