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 \[ d=\frac{\left\vert (1)(2)+(4)(3)+(1)(-1)-5\right\vert }{\sqrt{1+16+1}}= \frac{8}{3\sqrt{2}}=\frac{4\sqrt{2}}{3} \]