\begin{array}{l}\text{If the distance of the point}(4,6,8)\text{from the plane }\overrightarrow{r}.(6\hat{i}-12\hat{j}+4\hat{k})=a\text{ is}\mathrm{ 1},\text{then}\mathrm{ a }\text{is:}\end{array}

The correct option is: -2 or -30

To find the value of $a$, we start by converting the vector equation of the plane into its Cartesian form.
The given vector equation of the plane is:

$$\overrightarrow{r}·(6\hat{i}-12\hat{j}+4\hat{k})=a$$

By substituting the position vector $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$, we obtain the Cartesian equation of the plane:
$6x-12y+4z=a$, which can be rewritten as:

$$6x-12y+4z-a=0$$

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

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

Given details from the question:
• Point $(x_1,y_1,z_1)=(4,6,8)$
• Distance $d=1$
��� Plane parameters: $A=6$, $B=-12$, $C=4$, and $D=-a$

Substituting these values into the distance formula gives:

$$1=\frac{\left|6\right(4)-12(6)+4(8)-a|}{\sqrt{6^2+{(-12)}^2+4^2}}$$

Simplifying the expression under the square root and inside the absolute value:

$$1=\frac{|24-72+32-a|}{\sqrt{36+144+16}}$$

$$1=\frac{|-16-a|}{\sqrt{196}}$$

Since $\sqrt{196}=14$ and $|-16-a|=|a+16|$, we have:

$$1=\frac{|a+16|}{14}$$

Multiplying both sides by 14 yields:

$$|a+16|=14$$

This absolute value equation splits into two cases:

Case 1:
$$a+16=14$$ $$a=14-16$$ $$a=-2$$

Case 2:
$$a+16=-14$$ $$a=-14-16$$ $$a=-30$$

Thus, the possible values for $a$ are -2 or -30.