If the line joining the points $A(1,0,0)$ and $B(0,0,1)$ is a normal to the plane $\pi$ which passes through the point $A$, then the angle between the planes $\pi$ and $x+y+z=6$ is

The value of $m$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{2z-m}{3}$ lies in the plane $2x-5y+2z=7$,is

Let $P$ be the plane passing through the intersection of the planes $\overrightarrow{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\overrightarrow{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$,and the point $(2, 1, -2)$. Let the position vectors of the points $X$ and $Y$ be $\hat{i} - 2\hat{j} + 4\hat{k}$ and $5\hat{i} - \hat{j} + 2\hat{k}$ respectively. Then the points:

Let $S$ be the reflection of a point $Q$ with respect to the plane given by $\vec{r} = -(t+p) \hat{i} + \hat{j} + (1+p) \hat{k}$,where $t, p$ are real parameters and $\hat{i}, \hat{j}, \hat{k}$ are the unit vectors along the three positive coordinate axes. If the position vectors of $Q$ and $S$ are $10 \hat{i} + 15 \hat{j} + 20 \hat{k}$ and $\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ respectively,then which of the following is/are $TRUE$?
$(A)$ $3(\alpha+\beta) = -101$
$(B)$ $3(\beta+\gamma) = -71$
$(C)$ $3(\gamma+\alpha) = -86$
$(D)$ $3(\alpha+\beta+\gamma) = -121$

Find the equation of the plane which is perpendicular to the plane $5x + 3y + 6z + 8 = 0$ and which contains the line of intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 = 0$.

If $P(3, 2, 6)$ is a point in space and $Q$ is 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: