A variable straight line always intersects the lines $x=c,y=0;\,\,y=c,z=0;\,\,z=c,x=0.\quad$Find the equation of its locus.
A plane passing through first line is $x-c+\lambda1y=0$ and a plane passing through second line is$y-c+\lambda2z=0$ and these two planes taken together give the variable straight line.
Now I am not getting any idea on how to proceed with finding the locus.
Appreciate any help.
I might proceed as follows:
Consider a variable (finite) point $\mathbf p(\lambda)=(c,0,\lambda)$ on the first line. Together with the second line, this defines a plane. Switching to homogeneous coordinates, this plane is a null vector of the matrix $$\begin{bmatrix}0&c&0&1\\1&0&0&0\\c&0&\lambda&1\end{bmatrix},$$ which has as its first two rows a pair of points on the second line. Computing its null space produces the plane $\mathbf\pi(\lambda)=[0:\lambda:c:-c\lambda]$. The intersection of this plane with the third line is a null vector of $$\begin{bmatrix}0&0&1&-c\\1&0&0&0\\0&\lambda&c&-c\lambda\end{bmatrix},$$ i.e., the intersection of $\mathbf\pi$ with two planes that contain the third line. This yields for the homogeneous coordinates of the intersection point $[0:c\lambda-c^2:c\lambda:\lambda]$, or in inhomogeneous Cartesian coordinates, $\mathbf q(\lambda)=\left(0,{c\lambda-c^2\over\lambda},c\right)$. ($\mathbf\pi(0)$ is parallel to the third line, so it makes sense that you’d get a point at infinity for that value of the parameter.)
With these two points you have a parameterization of the ruled surface: $\mathbf r(\lambda,\mu) = (1-\mu)\mathbf p(\lambda)+\mu\mathbf q(\lambda)$. Setting this equal to $(x,y,z)$ and eliminating the two parameters should give you a Cartesian equation of the surface, which turns out to be a hyperboloid of one sheet with axis in the direction of $(1,1,1)$, just as achille hui suggests in his comment.
Let $L$ be the collection of lines in $\mathbb{R}^3$. For any $\ell \in L$, we can describe it by picking a point $p \in \ell$ and specify the "direction" of $\ell$ using a tangent vector $t$. The line $\ell$ will be the set of points obtainable from $p$ by translation along the $\pm t$ directions.
$$\ell = p + \mathbb{R}t \stackrel{def}{=} \{ p + \lambda t : \lambda \in \mathbb{R} \}$$
For any line $\ell = p + \mathbb{R}t$ and point $q \not\in \ell$, there exists a unique plane containing $q$ and $\ell$. We will denote this plane as $P(q,\ell)$. It is easy to see for any generic point $r \in \mathbb{R}^3$, $r$ belongs to $P(q,\ell)$ when and only when following equation is satisfied:
$$(r - q) \cdot (p - q)\times t = 0$$
Given three skew lines $ \ell_1 = p_1 + \mathbb{R}t_1, \ell_2 = p_2 + \mathbb{R}t_2, \ell_3 = p_3 + \mathbb{R}t_3 $ in general positions (i.e. pairwise disjoint and non-parallel to each other). Let $X$ be the locus of lines intersecting all three skew lines. More precisely,
$$X = \bigcup \big\{ \ell \in L : \ell \cap \ell_1 \ne \emptyset, \ell \cap \ell_2 \ne \emptyset, \ell \cap \ell_3 \ne \emptyset \big\}$$
Notice for any $q \notin \ell_2\cup\ell_3$, the planes $P(q,\ell_2)$ and $P(q,\ell_3)$ are defined. Since $\ell_2$ and $\ell_3$ are disjoint and not-parallel to each other, they cannot be contained in a single plane. This means $P(q,\ell_2) \ne P(q,\ell_3)$. Since these two planes intersect at $q$, they intersect at a line. In short, for any $q \notin \ell_2\cup\ell_3$, there is always a line passing through $q$ which intersect $\ell_2$ and $\ell_3$.
Apply these to point $q$ on $\ell_1$, we find there is always a line $\ell \in L$ which contains $q$. As a result, $\ell_1 \subset X$. By a similar argument, we have $\ell_2, \ell_3 \subset X$.
For point $q \notin \ell_1\cup\ell_2\cup\ell_3$, we have following $3$ lines through $q$. $$P(q,\ell_2)\cup P(q,\ell_3),\quad P(q,\ell_3)\cup P(q,\ell_1)\quad\text{ and }\quad P(q,\ell_1)\cup P(q,\ell_2)$$ In order for $q \in X$, this 3 lines need to coincide. There is equivalent to finding a direction $t$ which satisfy: $$ t \cdot (p_1 - q)\times t_1 = t \cdot (p_2 - q)\times t_2 = t \cdot (p_3 - q)\times t_3 = 0$$ We can recast this to matrix form $A t = 0$ where $A^T$ is a $3 \times 3$ matrix whose $i^{th}$ column is the column vector $(p_i - q) \times t_i$. The necessary and sufficient condition for a non-trivial solution of $t$ becomes
$$\det(A) = (( p_1 - q)\times t_1 ) \cdot \left[ (( p_2 - q)\times t_2 ) \times (( p_3 - q)\times t_3 )) \right] = 0$$
For the problem at hand, we can set
$$ q = (x,y,z) \;\text{ and }\; \begin{cases} p_1 = (c,0,0), t_1 = (0,0,1)\\ p_2 = (0,c,0), t_2 = (1,0,0)\\ p_3 = (0,0,c), t_3 = (0,1,0) \end{cases} \quad \longrightarrow \quad \begin{cases} (p_1 - q)\times t_1 = (-y,x-c,0)\\ (p_2 - q)\times t_2 = (0, -z,y-c)\\ (p_3 - q)\times t_3 = (z-c,0,-x) \end{cases} $$ The three lines in question do satisfy the condition we discussed before. In order for $q$ outside these lines belongs to $X$, the condition becomes:
$$\det\begin{bmatrix} -y & 0 & z-c\\ x-c & -z & 0\\ 0 & y-c & -x \end{bmatrix} = (x-c)(y-c)(z-c)-xyz = 0$$
With a little bit of algebra, we can rearrange RHS to the equation of the hyperboloid I mentioned in comment:
$$(x+y+z-c)^2 = x^2+y^2+z^2 - c^2$$
It is easy to see points on the three given lines also satisfy this equation. As a result, the locus $X$ we seek is the hyperboloid described by above equation.
