Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4ax-y+5z-7a=0=2x-5y-z-3, a \in R$ at the point $P(\alpha, \beta, \gamma)$. Then the value of $\alpha+\beta+\gamma$ equals...

For $a, b \in \mathbb{Z}$ and $|a - b| \leq 10$,let the angle between the plane $P: ax + y - z = b$ and the line $l: x - 1 = \frac{-y}{1} = z + 1$ be $\cos^{-1}\left(\frac{1}{3}\right)$. If the distance of the point $(6, -6, 4)$ from the plane $P$ is $3\sqrt{6}$,then $a^4 + b^2$ is equal to

Find the angle between the plane $\bar{r} \cdot (1, 2, 1) = 1$ and the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-1}$.

If the line joining the points $\overline{i} + 2\overline{j}$ and $\overline{j} - 2\overline{k}$ intersects the plane passing through the points $2\overline{i} - \overline{j}$,$2\overline{j} + 3\overline{k}$,and $\overline{k} - 2\overline{i}$ at $\overline{r}$,then $\overline{r} \cdot (\overline{i} + \overline{j} + \overline{k}) = $

Let a line with direction ratios $a, -4a, -7$ be perpendicular to the lines with direction ratios $3, -1, 2b$ and $b, a, -2$. If the point of intersection of the line $\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$ and the plane $x - y + z = 0$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to $.......$

The vector equation of the plane passing through the point $i + 2j - k$ and perpendicular to the line of intersection of the planes $r \cdot (3i - j + k) = 1$ and $r \cdot (i + 4j - 2k) = 2$ is