Answer
$\dfrac{2}{\sqrt 3}$ or, $\dfrac{2\sqrt 3}{3}$
Work Step by Step
Given: Equation of plane: $x+y+z=1$ and point $(2,0,-3)$
Need to apply Lagrange Multipliers Method to determine the shortest distance from the point $(2,0,-3)$ to the plane $x+y+z=1$.
We have $\nabla f=\lambda \nabla g$
Use the formula to calculate the shortest distance.
$D=|\dfrac{pa+qb+rc-s}{\sqrt{p^2+q^2+r^2}}|=|\dfrac{(1)(2)+(1)(0)+(1)(-3)-1}{\sqrt{1^2+1^2+1^2}}|$
Hence, $D=\dfrac{2}{\sqrt 3}$ or, $\dfrac{2\sqrt 3}{3}$