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

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors and $L$ be the line passing through the points $\vec{a}-\vec{b}+\vec{c}$ and $\vec{b}-\vec{c}$. If $\pi$ is a plane passing through the points $2\vec{a}-\vec{b}, 2\vec{b}-\vec{c}$ and $2\vec{c}-\vec{a}$, then the point of intersection of $L$ and $\pi$ is

Let $P(3, 2, 6)$ be a point in space and $Q$ be a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. Then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane $x + 2y + 3z = 15$ at a point $P$,then the distance of $P$ from the origin is

If the line $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$ is parallel to the plane $ax + by + cz + d = 0$,then:

The angle between the line $\frac{x - 2}{a} = \frac{y - 2}{b} = \frac{z - 2}{c}$ and the plane $ax + by + cz + 6 = 0$ is ......... $^o$