Find the vector equation of the plane which is at a distance of \frac{7}{\sqrt{38}} from the origin and the normal vector from origin is 2\hat{i}+3\hat{j}-5\hat{k}?

(a) \vec{r}.(\frac{2\hat{i}}{38}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{56}}

(b) \vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}

(c) \vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{5\hat{j}}{\sqrt{38}}+\frac{3\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}

(d) \vec{r}.(\frac{2\hat{i}}{\sqrt{58}}-\frac{3\hat{j}}{\sqrt{37}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}

The question was asked in exam.

My question is from Three Dimensional Geometry topic in chapter Three Dimensional Geometry of Mathematics – Class 12

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The correct answer is (b) \vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}

The best I can explain: Let \vec{n}=2\hat{i}+3\hat{j}-5\hat{k}

\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{(2\hat{i}+3\hat{j}-5\hat{k})}{\sqrt{(2^2+3^2+(-5)^2)}}=\frac{(2\hat{i}+3\hat{j}-5\hat{k})}{\sqrt{38}}

Hence, the required equation of the plane is \vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}