Find the equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to the plane $x-y+z=0$.

If the equation of the line passing through the point $(0, -\frac{1}{2}, 0)$ and perpendicular to the lines $\overrightarrow{r} = \lambda(\hat{i} + a\hat{j} + b\hat{k})$ and $\overrightarrow{r} = (\hat{i} - \hat{j} - 6\hat{k}) + \mu(-b\hat{i} + a\hat{j} + 5\hat{k})$ is $\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}$,then $a+b+c+d$ is equal to :

Find the equation of the plane containing the point $(0, 7, -7)$ and the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$.

The direction cosines of the line of intersection of the planes $x+2y+z-4=0$ and $2x-y+z-3=0$ are

Let $L$ be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If $L$ makes an angle $\alpha$ with the positive direction of the $x$-axis,then $\cos \alpha$ is:

Let $P$ be a plane $lx + my + nz = 0$ containing the line $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$. If plane $P$ divides the line segment $AB$ joining points $A(-3, -6, 1)$ and $B(2, 4, -3)$ in the ratio $k : 1$,then the value of $k$ is equal to