If $\overrightarrow{\mathrm{a}}=2\hat{\mathrm{i}}+\hat{\mathrm{j}}+3\hat{\mathrm{k}},\overrightarrow{\mathrm{b}}=3\hat{\mathrm{i}}+3\hat{\mathrm{j}}+\hat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{c}}={\mathrm{c}}_1\hat{\mathrm{i}}+{\mathrm{c}}_2\hat{\mathrm{j}}+{\mathrm{c}}_3\hat{\mathrm{k}}$ are coplanar vectors and $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{c}}=5,\overrightarrow{\mathrm{b}}\perp \overrightarrow{\mathrm{c}}$, then $122\left({\mathrm{c}}_1+{\mathrm{c}}_2+{\mathrm{c}}_3\right)$ is equal to _____________.

We are given three vectors:

• a → = 2î + ĵ + 3k̂
• b → = 3î + 3ĵ + k̂
• c → = c₁î + c₂ĵ + c₃k̂

Conditions:

1. The vectors are coplanar, meaning the scalar triple product of a →, b →, and c → is zero.
2. a → ⋅ c → = 5
3. b → ⊥ c →, i.e., the dot product b → ⋅ c → = 0.

Step 1: From the dot product condition, a → ⋅ c → = 5, we have:

2c₁ + c₂ + 3c₃ = 5    ...(i)

Step 2: From the orthogonality condition, b → ⋅ c → = 0, we get:

3c₁ + 3c₂ + c₃ = 0    ...(ii)

Step 3: Using the scalar triple product condition, we calculate:

[a → b → c →] = | 2  1  3 |
                 | 3  3  1 |
                 | c₁ c₂ c₃ |

Expanding the determinant:

[a → b → c →] = 2(3c₃ - c₂) - 1(3c₃ - c₁) + 3(3c₂ - 3c₁)

Simplifying:

[a → b → c →] = 3c₃ + 7c₂ - 8c₁ = 0    ...(iii)

Step 4: Solve the system of equations (i), (ii), and (iii):

1. From equations (i), (ii), and (iii), solve for c₁, c₂, c₃.
2. After solving, we get:
   • c₁ = 10/122
   • c₂ = -85/122
   • c₃ = 225/122

Step 5: Compute 122(c₁ + c₂ + c₃):

122(c₁ + c₂ + c₃) = 122(10/122 - 85/122 + 225/122)

Simplifying:

122(c₁ + c₂ + c₃) = 10 - 85 + 225 = 150

Final Answer: 150