If $\mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \mathbf{b}=-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \mathbf{c}=5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\mathbf{d}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ are four vectors, then $(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=$

$3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+5 \hat{\mathbf{j}}$ are the position vectors of three non-collinear points $A, B, C$ respectively. If the perpendicular drawn from $C$ onto $\mathbf{A B}$ meets $\mathbf{A B}$ at the point $a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$, then $a+b+c=$

Let $x_1, x_2, \ldots, x_{11}$ be the observations satisfying $\sum\limits_{i=1}^{11}\left(x_i-4\right)=22$ and $\sum\limits_{i=1}^{11}\left(x_i-4\right)^2=154$. If the mean and variance of the observations are $\alpha$ and $\beta$, then the quadratic equation having the roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is