The equation of the plane in which the lines $\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{-5}$ and $\frac{x - 8}{7} = \frac{y - 4}{1} = \frac{z - 5}{3}$ lie,is

The image of the point with position vector $\vec{i} + 3\vec{k}$ in the plane $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 1$ is

Suppose the line $\frac{x-2}{\alpha}=\frac{y-2}{-5}=\frac{z+2}{2}$ lies on the plane $x+3y-2z+\beta=0$. Then $(\alpha+\beta)$ is equal to ... .

Equation of the plane containing the line $x + 2y + 3z - 5 = 0 = 3x + 2y + z - 5$ which is parallel to the line $x - 1 = 2 - y = z - 3$ is:

Let $d$ be the distance between the foot of perpendiculars of the points $P(1, 2, -1)$ and $Q(2, -1, 3)$ on the plane $-x + y + z = 1$. Then $d^{2}$ is equal to

If the line joining the points $\hat{i}+\hat{j}$ and $3 \hat{i}+\hat{j}-\hat{k}$ meets the plane that passes through the point $2 \hat{i}+4 \hat{j}$ and is parallel to the vectors $3 \hat{j}+5 \hat{k}$ and $3 \hat{i}-\hat{k}$ at point $P$,then the position vector of the point $P$ is