The equation of a plane containing the line $x-2=\frac{y-4}{4}=\frac{z-6}{7}$ and parallel to the line $\vec{r}=(\hat{i}+3\hat{j}+5\hat{k})+\lambda(3\hat{i}+5\hat{j}+7\hat{k})$ is

Find the image of the point having position vector $\hat{i}+3 \hat{j}+4 \hat{k}$ in the plane $\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$.

If the equation of the plane containing the line $x+2y+3z-4=0=2x+y-z+5$ and perpendicular to the plane $\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2\hat{j}+3\hat{k})$ is $ax+by+cz=4$,then $(a-b+c)$ is equal to

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,then $(\alpha, \beta)=$

The distance of the point $(3, 8, 2)$ from the line $\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 2}{3}$ measured parallel to the plane $3x + 2y - 2z = 0$ is

$A$ line with direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. If the line meets the plane $2x + y + z = 9$ at point $Q$,then the length of $PQ$ is . . . . . . .