The line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is parallel to the plane

If the equation of the plane passing through the line of intersection of the planes $2x - y + z = 3$ and $4x - 3y + 5z + 9 = 0$ and parallel to the line $\frac{x + 1}{-2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $ax + by + cz + 6 = 0$,then $a + b + c$ is equal to $.............$.

If $4x + 4y - kz = 0$ is the equation of the plane through the origin that contains the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4},$ then $k =$

$A$ line with positive direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. If the line meets the plane $2x + y + z = 9$ at point $Q,$ then the length $PQ$ equals

Let $L$ be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If $L$ makes an angle $\alpha$ with the positive $X$-axis,then $\cos \alpha$ equals

If the lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s, s \in R$ and $x = \frac{t}{2}, y = 1 + t, z = 2 - t, t \in R$ are coplanar,then $\lambda = $