The value of $\lambda$ for which the straight line $\frac{x-\lambda}{3}=\frac{y-1}{2+\lambda}=\frac{z-3}{-1}$ may lie on the plane $x-2y=0$ is

If the line $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2x - 4y + z = 7$,then $k = \dots$

The equation of the plane that contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ and is perpendicular to the plane $5x+3y-6z+8=0$ is

Let $\pi_1$ be the plane determined by the vectors $\bar{i}+\bar{j}$ and $\bar{i}+\bar{k}$, and $\pi_2$ be the plane determined by the vectors $\bar{j}-\bar{k}$ and $\bar{k}-\bar{i}$. Let $\bar{a}$ be a non-zero vector parallel to the line of intersection of the planes $\pi_1$ and $\pi_2$. If $\bar{b}=\bar{i}+\bar{j}-\bar{k}$, then the angle between the vectors $\bar{a}$ and $\bar{b}$ is:

Let the image of the point $P(1, 6, a)$ in the line $L: \frac{x-1}{2} = \frac{y-2}{b} = \frac{z-a+1}{1}, b>0$, be $Q(\frac{a}{3}, 0, a+c)$. If $S(\alpha, \beta, \gamma), \alpha > 0$, is the point on $L$ such that the distance of $S$ from the foot of perpendicular $F$ from the point $P$ on $L$ is $2\sqrt{14}$, then $\alpha + \beta + \gamma$ is equal to:

If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to the planes $3x - 2y + z = 9$ and $x + y + z = 9$ is $x + by + cz + d = 0$, then $d =$