The area of the parallelogram with vertices $A(1, 2, 3)$,$B(1, 3, a)$,$C(3, 8, 6)$,and $D(3, 7, 3)$ is $\sqrt{265}$ sq. units. Then $a=$

Vectors $\vec{p}=a \hat{i}+b \hat{j}+c \hat{k}$, $\vec{q}=d \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}$ form a triangle $ABC$ such that $\vec{p}=\vec{q}+\vec{r}$. If the area of $\triangle ABC$ is $5 \sqrt{6}$ sq. units, then the sum of the absolute values of $a, b, c$ is

Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $\vec{a} = a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}$,$\vec{b} = b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$,and $\vec{c} = c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}$. Then show that $\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$.

Let the position vectors of two points $P$ and $Q$ be $3 \hat{i} - \hat{j} + 2 \hat{k}$ and $\hat{i} + 2 \hat{j} - 4 \hat{k}$ respectively. Let $R$ and $S$ be two points such that the direction ratios of lines $PR$ and $QS$ are $(4, -1, 2)$ and $(-2, 1, -2)$ respectively. Let lines $PR$ and $QS$ intersect at $T$. If the vector $\vec{TA}$ is perpendicular to both $\vec{PR}$ and $\vec{QS}$ and the length of vector $\vec{TA}$ is $\sqrt{5}$ units,then the modulus of a position vector of $A$ is

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$ and $\vec{a} \times \vec{c} = \vec{b}$, $\vec{a} \cdot \vec{c} = 3$, then $\vec{c}$ is

If the vertices of a triangle are $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$,then the area of the triangle is: