Find the equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z + 5 = 0$ and the point $(1, 1, 1)$.

The point of intersection of the line $\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{-2}$ and the plane $2x - y + 3z - 1 = 0$ is:

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

If the line joining the points $\hat{i}+\hat{j}$ and $3 \hat{i}+\hat{j}-\hat{k}$ meets the plane that passes through the point $2 \hat{i}+4 \hat{j}$ and is parallel to the vectors $3 \hat{j}+5 \hat{k}$ and $3 \hat{i}-\hat{k}$ at point $P$,then the position vector of the point $P$ is

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:

If a line $L$ is common to the planes $x-y+z+2=0$ and $2x+y-2z+5=0$,then the direction cosines of the line $L$ are