Find the vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2 \hat{i}-3 \hat{j}+4 \hat{k} .$ Also find its cartesian form.

(N/A) Let the normal vector be $\vec{n} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k} .$
The unit normal vector $\hat{n}$ is given by $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{2^2 + (-3)^2 + 4^2}} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{4 + 9 + 16}} = \frac{2 \hat{i} - 3 \hat{j} + 4 \hat{k}}{\sqrt{29}} .$
The vector equation of a plane at a distance $d$ from the origin is $\vec{r} \cdot \hat{n} = d .$
Substituting the values,we get $\vec{r} \cdot \left( \frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k} \right) = \frac{6}{\sqrt{29}} .$
To find the cartesian form,substitute $\vec{r} = x \hat{i} + y \hat{j} + z \hat{k} :$
$(x \hat{i} + y \hat{j} + z \hat{k}) \cdot \left( \frac{2}{\sqrt{29}} \hat{i} - \frac{3}{\sqrt{29}} \hat{j} + \frac{4}{\sqrt{29}} \hat{k} \right) = \frac{6}{\sqrt{29}} .$
This simplifies to $\frac{2x - 3y + 4z}{\sqrt{29}} = \frac{6}{\sqrt{29}} ,$ which gives $2x - 3y + 4z = 6 .$