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

Find the equation of the plane containing the lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - 2\hat{k})$.

If the mirror image of the point $(1, 3, 5)$ with respect to the plane $4x - 5y + 2z = 8$ is $(\alpha, \beta, \gamma)$,then $5(\alpha + \beta + \gamma)$ equals:

Let the image of the point $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ in the plane $x-2y+z-2=0$ be $P$. If the distance of the point $Q(6, -2, \alpha)$,where $\alpha > 0$,from $P$ is $13$,then $\alpha$ is equal to $...........$.

The equation of the plane containing the line $\frac{x+1}{2}=\frac{y+2}{1}=\frac{z-2}{3}$ and the point $(1,-1,3)$ is

The largest value of $a$,for which the perpendicular distance of the plane containing the lines $\vec{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}+a\hat{j}-\hat{k})$ and $\vec{r}=(\hat{i}+\hat{j})+\mu(-\hat{i}+\hat{j}-a\hat{k})$ from the point $(2,1,4)$ is $\sqrt{3}$,is...