The sine of the angle between the straight line $\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}$ and the plane $2x-2y+z=5$ is

The ratio in which the plane $x - 2y + 3z = 17$ divides the line segment joining the points $A(-2, 4, 7)$ and $B(3, -5, 8)$ is:

Three lines $L_1: \overrightarrow{r} = \lambda \hat{i}, \lambda \in R$,$L_2: \overrightarrow{r} = \hat{k} + \mu \hat{j}, \mu \in R$,and $L_3: \overrightarrow{r} = \hat{i} + \hat{j} + v\hat{k}, v \in R$ are given. For which point$(s)$ $Q$ on $L_2$ can we find a point $P$ on $L_1$ and a point $R$ on $L_3$ such that $P, Q,$ and $R$ are collinear?

The lines $\frac{x-0}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda}$ and $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{\lambda}$ are coplanar. If $p$ is the plane containing these lines,then which of the following points lies on the plane for all values of $\lambda$?

If the points $(1, 1, p)$ and $(-3, 0, 1)$ are equidistant from the plane $\vec{r} \cdot (3 \hat{i} + 4 \hat{j} - 12 \hat{k}) + 13 = 0$,then find the value of $p$.

The length and foot of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ are