If the plane $3x - 4y - kz = 7$ contains the line $\frac{1 - x}{-2} = \frac{y + 1}{3} = \frac{z}{4}$,find the value of $k$.

Let $Q$ be the mirror image of the point $P(1, 2, 1)$ with respect to the plane $x + 2y + 2z = 16$. Let $T$ be a plane passing through the point $Q$ and containing the line $\vec{r} = -\hat{k} + \lambda(\hat{i} + \hat{j} + 2\hat{k}), \lambda \in R$. Then,which of the following points lies on $T$?

The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2x + 3y - z = 5$ contains which one of the following points?

If $\lambda_1 < \lambda_2$ are two values of $\lambda$ such that the angle between the planes $P_1: \vec{r} \cdot (3 \hat{i} - 5 \hat{j} + \hat{k}) = 7$ and $P_2: \vec{r} \cdot (\lambda \hat{i} + \hat{j} - 3 \hat{k}) = 9$ is $\sin^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$,then the square of the length of the perpendicular from the point $(38 \lambda_1, 10 \lambda_2, 2)$ to the plane $P_1$ is $...........$.

The equation of the plane passing through the line $\frac{x - 1}{5} = \frac{y + 2}{6} = \frac{z - 3}{4}$ and the point $(4, 3, 7)$ is

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