The position vectors of the point $A, B$ are $\mathbf{a}, \mathbf{b}$ respectively. If the position vector of the point $C$ is $\frac{a}{2}+\frac{b}{3}$, then

If $|\mathbf{a}|=1,|\mathbf{b}|=2,|\mathbf{a}-\mathbf{b}|^2+|\mathbf{a}+2 \mathbf{b}|^2=20$, then $(a, b)=$

Let $A B C$ be a triangle and $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be the position vectors of $A, B$ and $C$, respectively. Let $D$ divides $B C$ in the ratio $3: 1$ internally and $E$ divides $A D$ in the ratio $4: 1$ internally. Let $B E$ meet $A C$ in $F$. If $E$ divides $B F$ in the ratio $3: 2$ internally, then the position vector of $F$ is

If $\alpha, \beta$ and $\gamma$ are real numebrs such that

$$ \begin{aligned} & \left(\frac{7}{3}+\beta\right) \hat{\mathbf{i}}-\hat{\mathbf{j}}+(\alpha+\gamma) \hat{\mathbf{k}} \\ & =\frac{5}{3}(\alpha \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\beta(2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+(\hat{\mathbf{i}}+\gamma \hat{\mathbf{j}}+3 \hat{\mathbf{k}}), \text { then } \\ & 5 \alpha-9 \beta+13 \gamma= \end{aligned} $$