The point of intersection of the line $x+1=\frac{y+3}{3}=\frac{-z+2}{2}$ with the plane $3x+4y+5z=10$ is

For a positive real number $p$,if the perpendicular distance from a point $-\hat{i} + p\hat{j} - 3\hat{k}$ to the plane $\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 7$ is $6$ units,then $p=$

Let $L$ be the line of intersection of planes $\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=2$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=2$. If $P(\alpha, \beta, \gamma)$ is the foot of perpendicular on $L$ from the point $(1,2,0)$,then the value of $35(\alpha+\beta+\gamma)$ is equal to :

Find the equation of the plane which passes through the points $(0,1,2)$ and $(-1,0,3)$ and is perpendicular to the plane $2x+3y+z=5$.

The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1, 1, 0)$ is

The equation of the plane passing through the origin and containing the line $\frac{x - 1}{5} = \frac{y - 2}{4} = \frac{z - 3}{5}$ is: