The length of the ⊥ᵉʳ from the point (0, – 1, 3) to the plane 2x + y – 2z + 1 = 0 is

The correct option is 2.

To find the length of the perpendicular from a point to a plane, we can use the standard perpendicular distance formula.

The perpendicular distance $d$ from a point $(x_1,y_1,z_1)$ to a plane given by the equation $Ax+By+Cz+D=0$ is expressed as:

$$d=\frac{|A{x}_1+B{y}_1+C{z}_1+D|}{\sqrt{A^2+B^2+C^2}}$$

From the given problem, we have:
Point $(x_1,y_1,z_1)=(0,\&\#x2212;1,3)$
Plane equation: $2x+y\&\#x2212;2z+1=0$, where $A=2$, $B=1$, $C=\&\#x2212;2$, and $D=1$.

Now, let's substitute these values into the distance formula:

$$d=\frac{\left|2\right(0)+1(\&\#x2212;1)+(\&\#x2212;2\left)\right(3)+1|}{\sqrt{2^2+1^2+{(\&\#x2212;2)}^2}}$$

Simplify the terms inside the absolute value in the numerator:
$2\left(0\right)+1(\&\#x2212;1)+(\&\#x2212;2)\left(3\right)+1=0\&\#x2212;1\&\#x2212;6+1=\&\#x2212;6$

Taking the absolute value of the numerator:
$|\&\#x2212;6|=6$

Simplify the terms inside the square root in the denominator:
$2^2+1^2+{(\&\#x2212;2)}^2=4+1+4=9$

Taking the square root of the denominator:
$\sqrt{9}=3$

Substitute these simplified values back into the distance equation:

$$d=\frac{6}{3}=2$$

Thus, the length of the perpendicular from the point to the plane is 2.