If the line $\frac{x - 1}{2} = \frac{y + \alpha}{\alpha} = \frac{z - \beta}{2}$ lies in the plane $2x + y + z = 5$,then $\alpha + \beta$ is

If $P(2, \beta, \alpha)$ lies on the plane $x+2y-z-2=0$ and $Q(\alpha, -1, \beta)$ lies on the plane $2x-y+3z+6=0$,then the direction cosines of the line $PQ$ are:

The equation of the plane passing through the line of intersection of planes $\pi_1: 2x + 6y + 4z - 7 = 0$ and $\pi_2: x - y - 2z - 2 = 0$,and perpendicular to the plane $x + y + 2z - 5 = 0$ is:

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 = $

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors and $L$ be the line passing through the points $\vec{a}-\vec{b}+\vec{c}$ and $\vec{b}-\vec{c}$. If $\pi$ is a plane passing through the points $2\vec{a}-\vec{b}, 2\vec{b}-\vec{c}$ and $2\vec{c}-\vec{a}$,then the point of intersection of $L$ and $\pi$ is

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$,and passing through the point $\hat{i} + 2\hat{j} - \hat{k}$.