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The distance of plane r⃗⋅(6i^−3j^−2k^)+1=0\vec{r}\cdot(6\hat{i} - 3\hat{j} - 2\hat{k}) + 1 = 0 from origin is :

Solution

✅ Correct Option: 2

Distance =∣d∣∣n⃗∣=∣−1∣36+9+4=149=17= \frac{|d|}{|\vec{n}|} = \frac{|-1|}{\sqrt{36+9+4}} = \frac{1}{\sqrt{49}} = \frac{1}{7}.

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