Let a plane $P$ contain two lines $\overrightarrow{r} = \hat{i} + \lambda(\hat{i} + \hat{j}), \lambda \in R$ and $\overrightarrow{r} = -\hat{j} + \mu(\hat{j} - \hat{k}), \mu \in R$. If $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from the point $M(1, 0, 1)$ to $P$,then $3(\alpha + \beta + \gamma)$ equals

For real numbers $\alpha$ and $\beta \neq 0$,if the point of intersection of the straight lines $\frac{x-\alpha}{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta}=\frac{y-6}{3}=\frac{z-7}{3}$ lies on the plane $x+2y-z=8$,then $\alpha-\beta$ is equal to:

The value of $k$ such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies in the plane $2x-4y+z=7$ is:

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^o$ about its line of intersection with the plane $5x - 4y - 2z + 1 = 0$. The equation of the plane in its new position is:

$A$ plane $\pi_1$ contains the vectors $\bar{i}+\bar{j}$ and $\bar{i}+2\bar{j}$. Another plane $\pi_2$ contains the vectors $2\bar{i}-\bar{j}$ and $3\bar{i}+2\bar{k}$. $\bar{a}$ is a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If the angle $\theta$ between $\bar{a}$ and $\bar{i}-2\bar{j}+2\bar{k}$ is acute, then $\theta=$

The coordinates of the foot of the perpendicular from the point $(1, -2, 1)$ on the plane containing the lines $\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}$ and $\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}$ is