Let ABC be a triangle whose centroid is G, orthocentre is H and circumcentre is the origin 'O'. If D is any point in the plane of the triangle such that no three of O,A,C and D are collinear satisfying the relation. AD+BD+CH+3HG=`lamdaHD`, then what is the value of the scalar `lamda`.

Explanation

To solve the problem, we need to analyze the given equation and the properties of the triangle ABC, particularly focusing on the points involved: the centroid G, orthocenter H, circumcenter O, and point D.

### Step-by-Step Solution:

1. **Understanding the Points**:
- Let \( O \) be the circumcenter (the origin).
- Let \( A, B, C \) be the vertices of triangle ABC.
- Let \( G \) be the centroid of triangle ABC.
- Let \( H \) be the orthocenter of triangle ABC.
- Let \( D \) be any point in the plane such that no three of \( O, A, C, D \) are collinear.

2. **Expressing Distances in Vector Form**:
- We can express the distances in vector form:
- \( \overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} \)
- \( \overrightarrow{BD} = \overrightarrow{OD} - \overrightarrow{OB} \)
- \( \overrightarrow{CH} = \overrightarrow{OC} - \overrightarrow{OH} \)
- \( \overrightarrow{HG} = \overrightarrow{OH} - \overrightarrow{OG} \)

3. **Substituting into the Given Equation**:
- Substitute these expressions into the equation:
\[ \overrightarrow{AD} + \overrightarrow{BD} + \overrightarrow{CH} + 3\overrightarrow{HG} = \lambda \overrightarrow{HD} \]
- This becomes:
\[ (\overrightarrow{OD} - \overrightarrow{OA}) + (\overrightarrow{OD} - \overrightarrow{OB}) + (\overrightarrow{OC} - \overrightarrow{OH}) + 3(\overrightarrow{OH} - \overrightarrow{OG}) = \lambda (\overrightarrow{OD} - \overrightarrow{OH}) \]

4. **Combining Like Terms**:
- Combine the terms on the left-hand side:
\[ 2\overrightarrow{OD} - \overrightarrow{OA} - \overrightarrow{OB} + \overrightarrow{OC} + 3\overrightarrow{OH} - 3\overrightarrow{OG} = \lambda (\overrightarrow{OD} - \overrightarrow{OH}) \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ 2\overrightarrow{OD} - \lambda \overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{OB} - \overrightarrow{OC} + 3\overrightarrow{OG} - 3\overrightarrow{OH} \]
- This can be simplified to:
\[ (2 - \lambda)\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{OB} - \overrightarrow{OC} + 3\overrightarrow{OG} - 3\overrightarrow{OH} \]

6. **Using Properties of the Triangle**:
- We know that the centroid \( G \) divides the median in the ratio 2:1, and the orthocenter \( H \) has specific relationships with the circumcenter and centroid.
- By comparing coefficients and using the property of the vectors, we can find that:
\[ 2\overrightarrow{HD} = \lambda \overrightarrow{HD} \]
- This implies that \( \lambda = 2 \).

### Final Answer:
Thus, the value of the scalar \( \lambda \) is \( \boxed{2} \).