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In the question, position vectors of A and C are given as:

$3\hat i + 3\hat j + 5\hat k$ and $\hat i - 5\hat j - 5\hat k$.

As we know that, in any parallelogram the midpoints of both the diagonals are the same. Thus, given M as the midpoint of DB will imply that M will also be the midpoint of AC. Midpoint is the mean value of two vectors.

So,

\[

O\vec M = \dfrac{{O\vec A + O\vec C}}{2} \\

= \dfrac{{(3\hat i + 3\hat j + 5\hat k) + \hat i - 5\hat j - 5\hat k}}{2} \\

= 2\hat i - \hat j \\

\]

So, the position vector of M is $2\hat i - \hat j$.

Now, we will obtain the magnitude of projection of vector OM on vector C, by dividing the magnitude of the dot product of vector Om and vector OC by magnitude of vector OC.

Thus, Magnitude of the projection is,

\[

\dfrac{{\left| {O\vec M.O\vec C} \right|}}{{\left| {O\vec C} \right|}} \\

= \dfrac{{\left| {2 + 5} \right|}}{{\left| {\sqrt {1 + 25 + 25} } \right|}} \\

= \dfrac{7}{{\sqrt {51} }} \\

\]

In the above expression we found the magnitude of OC as $\sqrt {51} $(=\[\sqrt {1 + 25 + 25} \]). Also, for the projection, the angle between the vectors will be zero.

The magnitude of the projection will be $\dfrac{7}{{\sqrt {51} }}$.