The line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the curve $xy = c^2, z = 0$ if $c$ is equal to

Let $L_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$ and $L_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$,then $|5\alpha-11\beta-8\gamma|$ equals :

The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is:

Let $N$ be the foot of the perpendicular from the point $P(1, -2, 3)$ on the line passing through the points $(4, 5, 8)$ and $(1, -7, 5)$. Then the distance of $N$ from the plane $2x - 2y + z + 5 = 0$ is $.......$.

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 angle between the plane $x + y + z = 5$ and the line of intersection of the planes $3x + 4y + z - 1 = 0$ and $5x + 8y + 2z + 14 = 0$ is