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 of intersection of the planes $ax + by = 3$ and $ax + by + cz = 0$ $(a > 0)$ makes an angle $30^{\circ}$ with the plane $y - z + 2 = 0$,then the direction cosines of the line are:

Let $\bar{A}$ be a vector parallel to the line of intersection of planes $P_1$ and $P_2$ passing through the origin. $P_1$ is parallel to the vectors $2 \hat{j}+3 \hat{k}$ and $4 \hat{j}-3 \hat{k}$,and $P_2$ is parallel to $\hat{j}-\hat{k}$ and $3 \hat{i}+3 \hat{j}$. Then the angle between $\bar{A}$ and $2 \hat{i}+\hat{j}-2 \hat{k}$ is

The plane passing through the intersection of the planes $x + y + z = 1$ and $2x + 3y + z - 4 = 0$ and parallel to the $y$-axis also passes through the point:

Let a line $l$ pass through the origin and be perpendicular to the lines $l_1: \overrightarrow{r} = (\hat{i} - 11\hat{j} - 7\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ and $l_2: \overrightarrow{r} = (-\hat{i} + \hat{k}) + \mu(2\hat{i} + 2\hat{j} + \hat{k})$. If $P$ is the point of intersection of $l$ and $l_1$,and $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular from $P$ on $l_2$,then $9(\alpha + \beta + \gamma)$ is equal to:

Let $P$ be the plane,which contains the line of intersection of the planes $x + y + z - 6 = 0$ and $2x + 3y + z + 5 = 0$ and it is perpendicular to the $xy$-plane. Then the distance of the point $(0, 0, 256)$ from $P$ is equal to