If the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}$ and $\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}$ intersect at the point $P$,then the distance of the point $P$ from the plane $z = a$ is:

If the angle $\theta$ between the line $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ and the plane $2x - y + \sqrt{\lambda} z + 4 = 0$ is such that $\sin \theta = \frac{1}{3}$,the value of $\lambda$ is

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$,and passing through the point $\hat{i} + 2\hat{j} - \hat{k}$.

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$,where $\alpha > 0$,is $5$ units,then the foot of the perpendicular from $P$ to the plane is:

The line of shortest distance between the lines $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$ makes an angle of $\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$ with the plane $P: ax-y-z=0$,$(a>0)$. If the image of the point $(1,1,-5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta-\gamma$ is equal to $........$

The length (in units) of the projection of the line segment,joining the points $(5,-1,4)$ and $(4,-1,3)$,on the plane $x+y+z=7$ is