91Ó°ÊÓ

The given point is $\left(-1,5,3\right)$and the constraint is$x+2y-3z=4.$

It is given that the point is $\left(-1,5,3\right),$so the distance of a point from the given point is$\sqrt{(x+1{)}^2+(y−5{)}^2+(z−3{)}^2}.$

Let the function be the square of the above distance $D(x,y,z)=(x+1{)}^2+(y−5{)}^2+(z−3{)}^2$and the constraint as$g(x,y,z)=x+2y−3z−4.$

By using the Lagrange's method, the gradient of the above functions are,

$$\begin{gathered} \mathrm{∇}D(x,y,z)=2(x+1)\mathbf{i}+2(y−5)\mathbf{j}+2(z−3)\mathbf{k} \\ \mathrm{∇}\mathrm{g}(\mathrm{x},\mathrm{y},\mathrm{z})=\mathbf{i}+2\mathbf{j}−3\mathbf{k} \end{gathered}$$

Now, the system of the equation is

$$\begin{gathered} \mathrm{∇}D(x,y,z)=\mathrm{λ}\mathrm{∇}g(x,y,z) \\ 2(x+1)\mathbf{i}+2(y−5)\mathbf{j}+2(z−3)\mathbf{k}=\mathrm{λ}(\mathbf{i}+2\mathbf{j}−3\mathbf{k}) \end{gathered}$$

Here,$2(x+1)=\mathrm{λ},2(y−5)=2\mathrm{λ}\text{and}2(z−3)=−3\mathrm{λ}.$

The values of $\mathrm{λ},$we get are$\frac{2(x+1)}{1}=\frac{2(y−5)}{2}=\frac{2(z−3)}{−3}=\mathrm{λ}.$

So, $y=5+2(x+1)\text{and}z=3−3(x+1).$

Substitute the given values in the given constraint,

$$\begin{gathered} x+2y−3z=4 \\ x+2(5+2(x+1\left)\right)−3(3−3(x+1\left)\right)=4 \\ x+4x+9x+14=4 \\ x=−\frac{5}{7} \end{gathered}$$

Thus, the points where the extreme value may appear are $\left(−\frac{5}{7},\frac{39}{7},\frac{15}{7}\right).$

Hence the points on the given surface closest to$\left(-1,5,3\right)\mathrm{are}\left(−\frac{5}{7},\frac{39}{7},\frac{15}{7}\right).$