On which of the following lines does the point of intersection of the line $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ lie?

The equation of the plane passing through the line of intersection of the planes $\overline{r} \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k})=1$ and $\overline{r} \cdot(\hat{i}-\hat{j})+4=0$,and perpendicular to the plane $\overline{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+8=0$,is given by $\overline{r} \cdot(-5 \hat{i}+2 \hat{j}+12 \hat{k})=\mu$. Then the value of $\mu$ is:

The image of the point $(5, 2, 6)$ with respect to the plane $x + y + z = 9$ is

The acute angle between the line $\frac{x-5}{2}=\frac{y+1}{-1}=\frac{z+4}{1}$ and the plane $3x-4y-z+5=0$ is

The vector equation of the plane containing the lines $r=(\hat{i}+\hat{j})+t(\hat{i}+2 \hat{j}-\hat{k})$ and $r=(\hat{i}+\hat{j})+s(-\hat{i}+\hat{j}-2 \hat{k})$ is

Assertion $(A)$: The equation of the plane passing through the point $(4, 4, 4)$ and the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z = 0$ is $29x + 23y + 17z = 276$.
Reason $(R)$: The equation of the plane passing through the line of intersection of planes $P_1 = 0$ and $P_2 = 0$ is $P_1 + \lambda P_2 = 0, \lambda \in \mathbb{R}$.