If the orthocentre of the triangle whose vertices are $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}, 5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is $x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$, then

If the vectors $\mathbf{A B}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}, \mathbf{A C}=s \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$, $\mathbf{C B}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ from $\triangle A B C$, then the values of $p, q, r$ and $s$ such that the area of that $\triangle A B C$ is $5 \sqrt{6}$ are

Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three unit vectors such that $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=\frac{1}{\sqrt{2}}(\mathbf{b}+\mathbf{c})$ and $\mathbf{b}$ is not parallel to $\mathbf{c}$. If $\alpha$ and $\beta$ are the angles between $\mathbf{a}, \mathbf{b}$ and $\mathbf{a}, \mathbf{c}$ respectively then $\alpha-\beta=$

Let $\mathbf{O A}=\mathbf{a}, \mathbf{O B}=\mathbf{b}$ be two non collinear vectors,

$\mathbf{O P}=x_1 \mathbf{a}+y_1 \mathbf{b}, \mathbf{O Q}=x_2 \mathbf{a}+y_2 \mathbf{b}$ and $\mathbf{A}^{\prime} \mathbf{O}=\mathbf{O A}$,

$\mathbf{B}^{\prime} \mathbf{O}=\mathbf{O B}$. If $x_1=\frac{-3}{4}, x_2=\frac{1}{3}, y_1=\frac{7}{4}, y_2=\frac{5}{3}$, then