The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{0}$ is parallel to

If the three planes $x = 5$,$2x - 5ay + 3z - 2 = 0$,and $3bx + y - 3z = 0$ pass through a common line,then the value of $(a, b)$ is:

The angle between the line $\vec{r} = (\hat{i} + \hat{j} - 2\hat{k}) + \lambda (2\hat{i} - \hat{j} + \hat{k})$ and the normal to the plane $\vec{r} \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 2$ is:

If the straight lines $x = 1 + s, y = -3 - \lambda s, z = 1 + \lambda s$ and $x = t/2, y = 1 + t, z = 2 - t$,with parameters $s$ and $t$ respectively,are coplanar,then $\lambda$ equals

$L$ is a line passing through the point $A(1, 0, -3)$ and parallel to a line having direction ratios $0, 1, -2$. $P$ is a point on the line $L$ which is at a minimum distance from the plane $2x + 3y + 5z = 1$. Then, the equation of the plane through $P$ and perpendicular to $AP$ is

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 $T$?