The equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $3x+4y+5z=2$ and perpendicular to the $XY$-plane is

If $P=(0,1,0)$ and $Q=(0,0,1)$,then the length of the projection of the line segment $PQ$ on the plane $x+y+z=3$ is:

Let the equation of the plane passing through the line of intersection of the planes $x+2y+az=2$ and $x-y+z=3$ be $5x-11y+bz=6a-1$. For $c \in \mathbb{Z}$,if the distance of this plane from the point $(a, -c, c)$ is $\frac{2}{\sqrt{a}}$,then $\frac{a+b}{c}$ is equal to

The equation of a plane containing the line $x-2=\frac{y-4}{4}=\frac{z-6}{7}$ and parallel to the line $\vec{r}=(\hat{i}+3\hat{j}+5\hat{k})+\lambda(3\hat{i}+5\hat{j}+7\hat{k})$ is

The angle between the line $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}$ and the plane $\bar{r} \cdot(6 \hat{\imath}-2 \hat{\jmath}-3 \hat{k})=5$ is

Let a line $L$ passing through the point $P(1, 1, 1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line $L$ intersect the $yz$-plane at the point $Q$. Another line parallel to $L$ and passing through the point $S(1, 0, -1)$ intersects the $yz$-plane at the point $R$. Then the square of the area of the parallelogram $PQRS$ is equal to . . . . . . .