In a $\triangle A B C$, let $a, b, c, s, r, R, I, S, r_1, r_2, r_3$ stand for their usual meaning. Then Match the items of List-I with those of the items of List-II.
| List-I | List-II |
| A. |
I. |
| B. |
II. |
| C. |
III. |
| D. |
IV. |
| V. |
$$ \text { The correct match is } $$
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are the position vectors of the points $A, B, C$ respectively, then match the items of list-I with those of list-II.
| $$ \text { List-I } $$ |
$$ \text { List-II } $$ |
||
|---|---|---|---|
| A. | $$ \text { } \begin{aligned} \mathbf{a} & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, \\ \mathbf{b} & =3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \\ \mathbf{c} & =4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $$ |
I. | $A, B, C$ are collinear |
| B. | $$ \text { } \begin{aligned} \mathbf{a} & =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \\ \mathbf{b} & =3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}, \\ \mathbf{c} & =-3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \end{aligned} $$ |
II. | $\triangle A B C$ is an isosceles triangle |
| C. | $$ \begin{aligned} &\text { }\\ &\begin{aligned} & a=2 \hat{i}-\hat{j}+\hat{k}, \\ & b=\hat{i}-3 \hat{j}-5 \hat{k}, \\ & c=-3 \hat{i}-4 \hat{j}-4 \hat{k} \end{aligned} \end{aligned} $$ |
III. | $\triangle A B C$ is a right-angled triangle |
| D. | $$ \begin{aligned} & a=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \\ & b=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \\ & c=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \end{aligned} $$ |
IV. | $\triangle A B C$ is a right-angled isosceles triangle |
| V. | $$ \triangle A B C \text { is an equilateral triangle } $$ |
||
$$ \text { The correct match is } $$
If the point of intersection of the lines $\mathbf{r}=\hat{\mathbf{i}}-6 \hat{\mathbf{j}}+(p \sec \alpha) \hat{\mathbf{k}}+t(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ and $\mathbf{r}=4 \hat{\mathbf{j}}+\hat{\mathbf{k}}+\lambda(2 \hat{\mathbf{i}}+(p \tan \alpha) \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ is $8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}$, (where $\left.0<\alpha<\frac{\pi}{2}\right)$, then $p=$
$\mathbf{1}, \mathbf{m}, \mathbf{n}$ are three unit vectors in a right handed system and $L$ is a line through the points $A, B, C$ whose position vectors are $p \mathbf{1}+7 \mathbf{m}-6 \mathbf{n}, 2 \mathbf{1}+5 \mathbf{m}-4 \mathbf{n}$ and $1+4 \mathbf{m}-3 \mathbf{n}$ respectively. If the equation of the plane containing $L$ and the points ( $-p, p, p+1$ ) is $a x+b y+c z=1$, then $p(a+b+c)=$
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