Cross product

Let $P$ be the plane such that it contains the straight line $\dfrac{x-1}2=\dfrac{y-3}3=\dfrac{z+2}1$ and is perpendicular to the plane $x+2y+3z=4$. Let $P_1$ be the plane which passes through the point $(4,2,2)$ and is parallel to $P$.

Then which of the following statements is (are) TRUE ?

Let $\vec a,\vec b$ be two vectors, and let $P,Q$ and $R$ be the points with position vectors $\vec a$, $\vec b$ and $\vec a+\vec b$, respectively, with respect to the origin $O$. If $|\vec a+\vec b|=\sqrt{21}$, $|\vec a-\vec b|=3$, and $\vec a$ and $(\vec a-\vec b)$ are perpendicular to each other, then the area of the triangle $OPR$ is

Let a line $L$ passing through the point $(1,1,1)$ be perpendicular to both the vectors $2\hat i+2\hat j+\hat k$ and $\hat i+2\hat j+2\hat k$. If $P(a,b,c)$ is the foot of perpendicular from the origin on the line $L$, then the value of $34(a+b+c)$ is :

Let $\hat u$ and $\hat v$ be unit vectors inclined at an acute angle such that $\left|\hat u\times\hat v\right|=\dfrac{\sqrt3}2$. If $\vec A=\lambda\hat u+\hat v+\left(\hat u\times\hat v\right)$, then $\lambda$ is equal to:

Let $\vec a=\sqrt7\hat i+\hat j-\hat k$ and $\vec b=\hat j+2\hat k$. If $\vec r$ is a vector such that $\vec r\times\vec a+\vec a\times\vec b=\vec0$ and $\vec r\cdot\vec a=0$, then $\left|3\vec r\right|^2$ is equal to:

Let a line L be perpendicular to both the lines $L_1:\dfrac{x+1}3=\dfrac{y+3}5=\dfrac{z+5}7$ and $L_2:\dfrac{x-2}1=\dfrac{y-4}4=\dfrac{z-6}7$. If $\theta$ is the acute angle between the lines L and $L_3:\dfrac{x-\frac87}2=\dfrac{y-\frac47}1=\dfrac z2$, then $\tan\theta$ is equal to:

Let $\vec a=2\hat i+3\hat j+3\hat k$ and $\vec b=6\hat i+3\hat j+3\hat k$. Then the square of the area of the triangle with adjacent sides determined by the vectors $\left(2\vec a+3\vec b\right)$ and $\left(\vec a-\vec b\right)$ is :

Let the image of the point $P(0,-5,0)$ in the line $\dfrac{x-1}2=\dfrac y1=\dfrac{z+1}{-2}$ be the point R and the image of the point $Q\left(0,\dfrac{-1}2,0\right)$ in the line $\dfrac{x-1}{-1}=\dfrac{y+9}4=\dfrac{z+1}1$ be the point S. Then the square of the area of the parallelogram PQRS is ________.

Let $\vec a=4\hat i-\hat j+3\hat k$, $\vec b=10\hat i+2\hat j-\hat k$ and a vector $\vec c$ be such that $2\left(\vec a\times\vec b\right)+3\left(\vec b\times\vec c\right)=\vec0$. If $\vec a\cdot\vec c=15$, then $\vec c\cdot\left(\hat i+\hat j-3\hat k\right)$ is equal to :

Let a line $L_1$ pass through the origin and be perpendicular to the lines
$L_2:\vec r=(3+t)\hat i+(2t-1)\hat j+(2t+4)\hat k$ and
$L_3:\vec r=(3+2s)\hat i+(3+2s)\hat j+(2+s)\hat k$, $t,s\in\mathbf R$.
If $(a,b,c)$, $a\in\mathbf Z$, is the point on $L_3$ at a distance of $\sqrt{17}$ from the point of intersection of $L_1$ and $L_2$, then $(a+b+c)^2$ is equal to ________.

Let $\vec w = \hat i + \hat j - 2\hat k$, and $\vec u$ and $\vec v$ be two vectors, such that $\vec u \times \vec v = \vec w$ and $\vec v \times \vec w = \vec u$. Let $\alpha, \beta, \gamma$, and $t$ be real numbers such that

$$\vec u = \alpha\hat i + \beta\hat j + \gamma\hat k,\quad -t\alpha + \beta + \gamma = 0,\quad \alpha - t\beta + \gamma = 0,\quad\text{and}\quad \alpha + \beta - t\gamma = 0.$$

Match each entry in List-I to the correct entry in List-II and choose the correct option.

Let $\gamma \in \mathbb{R}$ be such that the lines $L_1 : \dfrac{x+11}{1} = \dfrac{y+21}{2} = \dfrac{z+29}{3}$ and $L_2 : \dfrac{x+16}{3} = \dfrac{y+11}{2} = \dfrac{z+4}{\gamma}$ intersect. Let $R_1$ be the point of intersection of $L_1$ and $L_2$. Let $O = (0, 0, 0)$, and $\hat{n}$ denote a unit normal vector to the plane containing both the lines $L_1$ and $L_2$.

Match each entry in List-I to the correct entry in List-II.

The correct option is

Let $\vec p = 2\hat i + \hat j + 3\hat k$ and $\vec q = \hat i - \hat j + \hat k$. If for some real numbers $\alpha$, $\beta$, and $\gamma$, we have

$$15\hat i + 10\hat j + 6\hat k = \alpha\left(2\vec p + \vec q\right) + \beta\left(\vec p - 2\vec q\right) + \gamma\left(\vec p \times \vec q\right),$$

then the value of $\gamma$ is _____.

Let $Q$ be the cube with the set of vertices $\{(x_1, x_2, x_3) \in \mathbb{R}^3 : x_1, x_2, x_3 \in \{0, 1\}\}$. Let $F$ be the set of all twelve lines containing the diagonals of the six faces of the cube $Q$. Let $S$ be the set of all four lines containing the main diagonals of the cube $Q$; for instance, the line passing through the vertices $(0,0,0)$ and $(1,1,1)$ is in $S$. For lines $\ell_1$ and $\ell_2$, let $d(\ell_1, \ell_2)$ denote the shortest distance between them. Then the maximum value of $d(\ell_1, \ell_2)$, as $\ell_1$ varies over $F$ and $\ell_2$ varies over $S$, is

Let the position vectors of the points $P, Q, R$ and $S$ be $\vec a = \hat i + 2\hat j - 5\hat k$, $\vec b = 3\hat i + 6\hat j + 3\hat k$, $\vec c = \dfrac{17}{5}\hat i + \dfrac{16}{5}\hat j + 7\hat k$ and $\vec d = 2\hat i + \hat j + \hat k$, respectively. Then which of the following statements is true?

Let $\hat i, \hat j$ and $\hat k$ be the unit vectors along the three positive coordinate axes. Let
$$\vec a = 3\hat i + \hat j - \hat k,\qquad \vec b = \hat i + b_2\hat j + b_3\hat k,\ b_2, b_3 \in \mathbb{R},\qquad \vec c = c_1\hat i + c_2\hat j + c_3\hat k,\ c_1, c_2, c_3 \in \mathbb{R}$$
be three vectors such that $b_2b_3 > 0$, $\vec a\cdot\vec b = 0$ and
$$\begin{pmatrix}0 & -c_3 & c_2\\ c_3 & 0 & -c_1\\ -c_2 & c_1 & 0\end{pmatrix}\begin{pmatrix}1\\ b_2\\ b_3\end{pmatrix} = \begin{pmatrix}3 - c_1\\ 1 - c_2\\ -1 - c_3\end{pmatrix}.$$
Then, which of the following is/are TRUE?

Let $\hat a,\hat b$ be unit vectors. If $\vec c$ be a vector such that the angle between $\hat a$ and $\vec c$ is $\dfrac\pi{12}$, and $\hat b=\vec c+2\left(\vec c\times\hat a\right)$, then $\left|6\vec c\right|^2$ is equal to :

If the shortest distance between the lines $\vec r=\left(-\hat i+3\hat k\right)+\lambda\left(\hat i-a\hat j\right)$ and $\vec r=\left(-\hat j+2\hat k\right)+\mu\left(\hat i-\hat j+\hat k\right)$ is $\sqrt{\dfrac23}$, then the integral value of $a$ is equal to ______.

If the shortest distance between the lines $\dfrac{x-1}2=\dfrac{y-2}3=\dfrac{z-3}\lambda$ and $\dfrac{x-2}1=\dfrac{y-4}4=\dfrac{z-5}5$ is $\dfrac1{\sqrt3}$, then the sum of all possible values of $\lambda$ is:

Let $\hat a$ and $\hat b$ be two unit vectors such that $\left|\left(\hat a+\hat b\right)+2\left(\hat a\times\hat b\right)\right|=2$. If $\theta\in(0,\pi)$ is the angle between $\hat a$ and $\hat b$, then among the statements:

(S1) : $2\left|\hat a\times\hat b\right|=\left|\hat a-\hat b\right|$

(S2) : The projection of $\hat a$ on $\left(\hat a+\hat b\right)$ is $\frac12$