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

The position vector of a point $P$ is $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\mathbf{a}=-\hat{\mathbf{i}}-2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ are two vectors which determine a plane $\pi$. The equation of a line through $P$ normal to $\mathbf{b}$ and lying on the plane $\pi$ is

In a quadrilateral $A B C D$, the point $P$ divides $D C$ in the ratio $1: 3$ internally and $Q$ is the mid-point of $A C$. If $\mathbf{A B}+\mathbf{A D}+\mathbf{B C}-2 \mathbf{D C}=\lambda \mathbf{P Q}$, then the value of $\lambda$ is

$\mathbf{p}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{q}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$. If the vectors $\mathbf{a}$ and $\mathbf{b}$ are the orthogonal projections of $\mathbf{p}$ on $\mathbf{q}$ and $\mathbf{q}$ on $\mathbf{p}$ respectively, then $\frac{\mathbf{a} \times \mathbf{b}}{\mathbf{a} \cdot \mathbf{b}}=$