Key Concepts and Formulas

• Dot Product: For vectors $\vec{u} = u_1\widehat i + u_2\widehat j + u_3\widehat k$ and $\vec{v} = v_1\widehat i + v_2\widehat j + v_3\widehat k$, the dot product is $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3$.
• Perpendicular Vectors: Two non-zero vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if $\vec{u} \cdot \vec{v} = 0$.
• Coplanar Vectors: Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if $\vec{c}$ can be expressed as a linear combination of $\vec{a}$ and $\vec{b}$ (provided $\vec{a}$ and $\vec{b}$ are not collinear), i.e., $\vec{c} = x\vec{a} + y\vec{b}$ for some scalars $x$ and $y$.

Step-by-Step Solution

Step 1: Express $\vec{c}$ using the coplanarity condition.

• Why: Since $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar, and $\vec{a}$ and $\vec{b}$ are not collinear (their components are not proportional), $\vec{c}$ can be written as a linear combination of $\vec{a}$ and $\vec{b}$. This approach simplifies the problem by introducing two unknowns ($x, y$) instead of three ($c_1, c_2, c_3$).
• Math: Let $\vec{c} = x\vec{a} + y\vec{b}$. $\vec{c} = x(2\widehat i + \widehat j + 3\widehat k) + y(3\widehat i + 3\widehat j + \widehat k)$ $\vec{c} = (2x+3y)\widehat i + (x+3y)\widehat j + (3x+y)\widehat k$ Comparing this with $\vec{c} = c_1\widehat i + c_2\widehat j + c_3\widehat k$, we get: $c_1 = 2x+3y$ $c_2 = x+3y$ $c_3 = 3x+y$

Step 2: Use the condition $\overrightarrow a \,.\,\overrightarrow c = 5$ to form an equation in $x$ and $y$.

• Why: This condition provides a linear equation involving the scalars $x$ and $y$ that define $\vec{c}$.
• Math: First, calculate the necessary dot products involving $\vec{a}$ and $\vec{b}$: $\vec{a} \cdot \vec{a} = (2)(2) + (1)(1) + (3)(3) = 4 + 1 + 9 = 14$ $\vec{a} \cdot \vec{b} = (2)(3) + (1)(3) + (3)(1) = 6 + 3 + 3 = 12$ Now, substitute $\vec{c} = x\vec{a} + y\vec{b}$ into the given condition: $\vec{a} \cdot (x\vec{a} + y\vec{b}) = 5$ $x(\vec{a} \cdot \vec{a}) + y(\vec{a} \cdot \vec{b}) = 5$ $14x + 12y = 5 \quad \text{(Equation 1)}$

Step 3: Use the condition $\overrightarrow b \bot \overrightarrow c$ to form another equation in $x$ and $y$.

• Why: The perpendicularity condition means their dot product is zero, giving us a second linear equation in $x$ and $y$.
• Math: Calculate the necessary dot products: $\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b} = 12$ $\vec{b} \cdot \vec{b} = (3)(3) + (3)(3) + (1)(1) = 9 + 9 + 1 = 19$ Substitute $\vec{c} = x\vec{a} + y\vec{b}$ into the given condition: $\vec{b} \cdot (x\vec{a} + y\vec{b}) = 0$ $x(\vec{b} \cdot \vec{a}) + y(\vec{b} \cdot \vec{b}) = 0$ $12x + 19y = 0 \quad \text{(Equation 2)}$

Step 4: Solve the system of linear equations for $x$ and $y$.

• Why: We have two independent equations (Equation 1 and Equation 2) with two unknowns ($x$ and $y$). Solving this system will give us the specific values of $x$ and $y$.
• Math: From Equation 2: $12x = -19y \implies x = -\frac{19}{12}y$. Substitute this into Equation 1: $14\left(-\frac{19}{12}y\right) + 12y = 5$ $-\frac{266}{12}y + 12y = 5$ $-\frac{133}{6}y + \frac{72}{6}y = 5$ $\frac{-133+72}{6}y = 5$ $-\frac{61}{6}y = 5 \implies y = -\frac{30}{61}$ Now substitute the value of $y$ back into the expression for $x$: $x = -\frac{19}{12}\left(-\frac{30}{61}\right) = \frac{19 \times 30}{12 \times 61} = \frac{19 \times 5}{2 \times 61} = \frac{95}{122}$ So, $x = \frac{95}{122}$ and $y = -\frac{30}{61}$.

Step 5: Calculate the sum $c_1 + c_2 + c_3$.

• Why: The problem asks for $122(c_1+c_2+c_3)$, so we need to find the sum $c_1+c_2+c_3$. We can do this by summing the expressions for $c_1, c_2, c_3$ in terms of $x$ and $y$.
• Math: $c_1 + c_2 + c_3 = (2x+3y) + (x+3y) + (3x+y)$ $c_1 + c_2 + c_3 = (2x+x+3x) + (3y+3y+y)$ $c_1 + c_2 + c_3 = 6x + 7y$ Substitute the calculated values of $x$ and $y$: $c_1 + c_2 + c_3 = 6\left(\frac{95}{122}\right) + 7\left(-\frac{30}{61}\right)$ $c_1 + c_2 + c_3 = \frac{6 \times 95}{122} - \frac{7 \times 30}{61}$ Since $122 = 2 \times 61$: $c_1 + c_2 + c_3 = \frac{3 \times 95}{61} - \frac{210}{61}$ $c_1 + c_2 + c_3 = \frac{285}{61} - \frac{210}{61} = \frac{285 - 210}{61} = \frac{75}{61}$

Step 6: Compute the final expression $122(c_1 + c_2 + c_3)$.

• Why: This is the final calculation required by the problem statement.
• Math: $122(c_1 + c_2 + c_3) = 122 \left(\frac{75}{61}\right)$ Since $122 = 2 \times 61$: $122 \left(\frac{75}{61}\right) = (2 \times 61) \left(\frac{75}{61}\right) = 2 \times 75 = 150$

Common Mistakes & Tips

• Scalar Triple Product vs. Linear Combination: While the scalar triple product $[\vec{a} \vec{b} \vec{c}] = 0$ is the definition of coplanarity, using $\vec{c} = x\vec{a} + y\vec{b}$ is often more practical when dot products are involved, as it leads directly to a system of linear equations for $x$ and $y$.
• Arithmetic Errors: Be extremely careful with fraction arithmetic and signs when solving the system of equations. Double-checking the calculations for $x$ and $y$ can save significant time.
• Commutativity of Dot Product: Remember that $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$, which simplifies some calculations.

Summary

The problem leverages the properties of coplanar and perpendicular vectors, along with the dot product. By expressing the unknown coplanar vector $\vec{c}$ as a linear combination of the given vectors $\vec{a}$ and $\vec{b}$, we transformed the geometric conditions into a system of two linear equations in two unknowns ($x$ and $y$). Solving this system allowed us to find the coefficients of the linear combination. Finally, we used these coefficients to calculate the sum of the components of $\vec{c}$ and then the required expression.

The final answer is $\boxed{150}$.