$$ABCD$$ is parallelogram. The position vectors of $$A$$ and $$C$$ are respectively, $$3\hat{i} + 3\hat{j} + 5\hat{k}$$ and $$\hat{i} - 5\hat{j} - 5\hat{k}$$. If $$M$$ is the midpoint of the diagonal $$DB$$, then the magnitude of the projection of $$\overrightarrow{OM}$$ on $$\overrightarrow{OC}$$, where $$O$$ is the origin, is

The position vectors of vertices are given from the origin $$O$$:
$$\overrightarrow{OA}=3\hat i+3\hat j+5\hat k$$, $$\overrightarrow{OC}=1\hat i-5\hat j-5\hat k$$.

In any parallelogram the diagonals bisect each other. Hence the midpoint $$M$$ of diagonal $$DB$$ is the same as the midpoint of diagonal $$AC$$.

Midpoint formula for vectors:
$$\overrightarrow{OM}= \dfrac{\overrightarrow{OA}+\overrightarrow{OC}}{2}$$

Compute $$\overrightarrow{OM}$$:
$$\overrightarrow{OM} =\dfrac{(3\hat i+3\hat j+5\hat k)+(1\hat i-5\hat j-5\hat k)}{2} =\dfrac{4\hat i-2\hat j+0\hat k}{2} =2\hat i-\hat j.$$

The required projection is the magnitude of the component of $$\overrightarrow{OM}$$ along $$\overrightarrow{OC}$$.
If two vectors are $$\mathbf{u}$$ and $$\mathbf{v}$$, the magnitude of the projection of $$\mathbf{u}$$ on $$\mathbf{v}$$ is $$\dfrac{|\mathbf{u}\cdot \mathbf{v}|}{|\mathbf{v}|}$$.

Dot product:
$$\overrightarrow{OM}\cdot\overrightarrow{OC} =(2\hat i-\hat j)\cdot(1\hat i-5\hat j-5\hat k) =2(1)+(-1)(-5)+0(-5)=2+5=7.$$

Magnitude of $$\overrightarrow{OC}$$:
$$|\overrightarrow{OC}|=\sqrt{1^2+(-5)^2+(-5)^2} =\sqrt{1+25+25}=\sqrt{51}.$$

Therefore, the required magnitude is
$$\dfrac{|\overrightarrow{OM}\cdot\overrightarrow{OC}|}{|\overrightarrow{OC}|} =\dfrac{7}{\sqrt{51}}.$$

Option D which is: $$\dfrac{7}{\sqrt{51}}$$