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 $a+b+c=$

If the vectors $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+l \hat{\mathbf{k}},-3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-4 l \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 / \hat{\mathbf{k}}$ form a right-angled triangle for a positive value of $l$, then the length of its hypotenuse is

A unit vector that is perpendicular to the vector $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and coplanar with the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ is