Equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and has the largest distance from the origin is

Let the equation of the plane $P$ containing the line $x+10=\frac{8-y}{2}=z$ be $ax+by+3z=2(a+b)$ and the distance of the plane $P$ from the point $(1,27,7)$ be $c$. Then $a^2+b^2+c^2$ is equal to $.............$.

If $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=-\hat{k}$ are position vectors of two points, and $\vec{b}=2 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{d}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors, then the lines $\vec{r}=\vec{a}+t \vec{b}$ and $\vec{r}=\vec{c}+s \vec{d}$ are:

Three lines $L_1: \overrightarrow{r} = \lambda \hat{i}, \lambda \in R$,$L_2: \overrightarrow{r} = \hat{k} + \mu \hat{j}, \mu \in R$,and $L_3: \overrightarrow{r} = \hat{i} + \hat{j} + v\hat{k}, v \in R$ are given. For which point$(s)$ $Q$ on $L_2$ can we find a point $P$ on $L_1$ and a point $R$ on $L_3$ such that $P, Q,$ and $R$ are collinear?

Let $A$ be a point on the line $\vec{r} = (1 - 3\mu)\hat{i} + (\mu - 1)\hat{j} + (2 + 5\mu)\hat{k}$ and $B(3, 2, 6)$ be a point in space. Then the value of $\mu$ for which the vector $\overrightarrow{AB}$ is parallel to the plane $x - 4y + 3z = 1$ is

If a plane $P$ passes through the points $(1,0,0)$ and $(0,1,0)$ and makes an angle $\frac{\pi}{4}$ with the plane $x+y=3$, then the direction ratios of a normal to that plane $P$ are