The projection of the line segment joining the points $P(1, -1, 3)$ and $Q(2, -4, 11)$ on the line joining the points $A(-1, 2, 3)$ and $B(3, -2, 10)$ is

The scalar product of vectors $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$ and a unit vector along the sum of vectors $\bar{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\bar{c}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}$ is $1$. Then the value of $\lambda$ is:

$AB=a$ and $AC=b$ are the sides of $\triangle ABC$. $P$ is a point on $AB$ and $Q$ is a point on $BC$ such that $\frac{AP}{PB}=\frac{1}{2}$ and $\frac{BQ}{QC}=\frac{1}{2}$. If the point of intersection of $AQ$ and $CP$ is $D$ and the area of $\triangle BCD$ is $7$ square units,then the area of the $\triangle ABC$ (in the same square units) is

If $a = (1, -1, 2)$,$b = (-2, 3, 5)$,$c = (2, -2, 4)$ and $i$ is the unit vector in the $x$-direction,then $(a - 2b + 3c) \cdot i = $

If the angle between the vectors $2 \alpha^2 \hat{i} + 4 \alpha \hat{j} + \hat{k}$ and $7 \hat{i} - 2 \hat{j} + \alpha \hat{k}$ is obtuse,then

Assertion $(A)$: $a, b, c, d$ are position vectors of $4$ points such that $2a - 3b + 7c - 6d = 0 \Rightarrow a, b, c, d$ are coplanar.
Reason $(R)$: Vector equation of the plane passing through three points whose position vectors are $a, b, c$ is $r = (1 - x - y)a + xb + yc$.
Which of the following is true?