The distance of the plane 2x – 3y + 6z + 7 = 0 from the point (2, -3, -1) is

The correct option is 2.

To find the distance of a plane from a given point, we use the standard formula for the perpendicular distance of a point $(x_1,y_1,z_1)$ from the plane $Ax+By+Cz+D=0$.

The formula for the distance $d$ is given by:
$$d=\frac{|A{x}_1+B{y}_1+C{z}_1+D|}{\sqrt{A^2+B^2+C^2}}$$

From the given equation of the plane, $2x-3y+6z+7=0$, we identify the coefficients:
$A=2$, $B=-3$, $C=6$, and $D=7$.

The given point is $(x_1,y_1,z_1)=(2,-3,-1)$.

Now, let's substitute these values into the distance formula step-by-step.

First, calculate the numerator:
$$\text{Numerator}=\left|2\right(2)+(-3\left)\right(-3)+6(-1)+7|$$
$$=|4+9-6+7|$$
$$=\left|14\right|=14$$

Next, calculate the denominator:
$$\text{Denominator}=\sqrt{2^2+{(-3)}^2+6^2}$$
$$=\sqrt{4+9+36}$$
$$=\sqrt{49}=7$$

Finally, divide the numerator by the denominator to find the distance $d$:
$$d=\frac{14}{7}=2$$

Thus, the distance of the plane from the given point is 2 units.