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The length of an edge of a cube $ABCDA_1B_1C_1D_1$is equal to unity. A point E taken on the edge AA$_1$ is such that AE = $\frac{1}{3}$ . A point F is taken on the edge BC such that BF = $\frac{1}{4}$. If O$_1$ is the centre of the cube,we have to find the shortest distance of the vertex B from the plane of the ∆ $O_1EF$.

I am not getting any start .

I know that equation of plane is $(\vec r -\vec a).\widehat n$ where $\widehat n $ is perpendicular vector of planen

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    That’s a good start. You’re given three points on the plane. How might you go about finding the plane’s perpendicular vector from them? Start with the simple case of $A$ at the origin and the opposite vertex at $(1,1,1)$. – 2017-01-21
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    @amd that is cross product of two vectors $O_1E$&$O_1F$ – 2017-01-23
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    Once you have that and any point on the plane, you can write down the normal equation of the plane $\mathbf n\cdot\mathbf x=\mathbf n\cdot\mathbf p$. From there, it’s just a matter of computing the distance between a point and a plane, which has been answered here many, many times. – 2017-01-23

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