$2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ are the position vectores of two points $A$ and $B$ respectively and $C$ divides $A B$ in the ratio $3: 2$ : If $3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ is the position of vector of a point $D$, then the unit vector in the direction of $C D$ is

A unit vector $\hat{\mathbf{e}}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$ is coplanar with the vectors $\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$, and $3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-5 \hat{\mathbf{k}}$. If $\hat{\mathbf{e}}$ is perpendicular to the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$, then $2 a^{2}+3 b^{2}+4 c^{2}=$

$\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \hat{\mathbf{b}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{c}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ are three vectors. If $\hat{\mathbf{d}}$ is a normal to the plane of $\hat{\mathbf{a}}$ and $\hat{\mathbf{b}}$ and d. $\hat{\mathbf{c}}=2$, then $|\hat{\mathbf{d}}|=$

If $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{c}=-\hat{\mathbf{k}}$ are position vectors of two points and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}, \mathbf{d}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ are two vectors, then the lines $\mathbf{r}=\mathbf{a}+t \mathbf{b}, \mathbf{r}=\mathbf{c}+s \mathbf{d}$ are