Let $L$ be the line of intersection of the planes $ax+by+cz=1$ and $a'x+b'y+c'z=1$.Find the perpendicular distance of $L$ from the origin. Let $L$ be the line of intersection of the planes $ax+by+cz=1$ and $a'x+b'y+c'z=1$.
Also,given that $$(a-a')^2+(b-b')^2+(c-c')^2=16$$ and $$(ab'-a'b)^2+(bc'-b'c)^2+(ac'-a'c)^2=9$$
Find the perpendicular distance of $L$ from the origin.
My Attempt
Let the direction ratios of the line $L$ be $l,m,n$ where
$al+bm+cn=0$
and $a'l+b'm+c'n=0$.
Solving for $l,m,n$
$l=bc'-b'c$
$m=a'c-ac'$
$n=ab'-a'b$
Let $P(\alpha,\beta,\gamma)$ be the foot of perpendicular dropped upon $L$ from the origin.
$a\alpha+b\beta+c \gamma=1$
$a'\alpha+b'\beta+c'\gamma=1$
which leads to $(a-a')\alpha+(b-b')\beta+(c-c')\gamma=0$
But beyond this I am not able to solve. How is one supposed to link all this information.
 A: Think of it as a constrained optimization problem.
$$\min x^2+y^2+z^2$$
s.t.
$$ax+by+cz=1$$
$$a'x+b'y+c'z=1$$
The Lagrangian becomes:
$$L = x^2+y^2+z^2-\lambda(ax+by+cz-1)-\mu(a'x+b'y+c'z-1)$$
Taking derivative with respect to $x$ and setting to $0$,
$$\hat{x}=\frac{a\lambda+a'\mu}{2}$$
and similarly for $\hat{y}$ and $\hat{z}$.
where the optimal point is $(\hat{x}, \hat{y}, \hat{z})$
Note that this is the same as the condition $\vec{c}=\lambda \vec{a}+\mu \vec{b}$ in the answer by @Maverick.
Now, the constraints must be satisfied, so
$$a\hat{x}+b\hat{y}+c\hat{z}=1$$
and
$$a'\hat{x}+b'\hat{y}+c'\hat{z}=1$$
Use these to get $\lambda$ and $\mu$.
A: Let us convert the question in vector form. It can be read as follows:
Let $L$ be line of intersection of planes $\vec r.\vec a=1$ and $\vec r.\vec b=1$, where $|\vec a-\vec b|=4$ and $|\vec a \times \vec b|=3$. Find perpendicular distance of the line $L$ from the origin.
Let $\vec c$ be the position vector of the foot of perpendicular dropped upon the line from origin. So we have to find $|\vec c|.$
Since the line is perpendicular to both $\vec a$ and $\vec b$,therefore the line must be parallel to $\vec a\times\vec b.$
Therefore,$\vec c$ is perpendicular to  $\vec a\times\vec b$.
Also $\vec c.\vec a=1$ and $\vec c.\vec b=1$ since $\vec c$ lies on $L$ which in turn lies on both the planes
So,  
$\vec c.(\vec a\times\vec b)=0$
$[\vec c\space   \vec a  \space \vec b]=0$
Thus $\vec c$ , $\vec a$ and $\vec b$ are coplanar and thus $\vec c$ can be expressed as a linear combination of $\vec a$ and $\vec b$.
$\vec c=\lambda \vec a+\mu \vec b$
Taking scalar(dot) product with $\vec c$ on both sides, we get
$\vec c.\vec c=\lambda \vec a.\vec c+\mu \vec b.\vec c$
$|\vec c|^2=\lambda+\mu$
Similarly, taking dot product with $\vec a$ and then by $\vec b$ on both sides we obtain the two equations
$\lambda a^2+\mu \vec a.\vec b=1$
$\lambda \vec a.\vec b+\mu b^2=1$
Solving for $\lambda$ and $\mu$ from the above two equations
$$\lambda=\frac{b^2-\vec a.\vec b}{a^2b^2-(\vec a.\vec b)^2}=\frac{b^2-\vec a.\vec b}{|\vec a\times\vec b|^2}$$
$$\mu=\frac{a^2-\vec a.\vec b}{a^2b^2-(\vec a.\vec b)^2}=\frac{a^2-\vec a.\vec b}{|\vec a\times\vec b|^2}$$
$$|\vec c|^2=\lambda+\mu=\frac{b^2-\vec a.\vec b}{|\vec a\times\vec b|^2}+\frac{a^2-\vec a.\vec b}{|\vec a\times\vec b|^2}=\frac{|\vec a-\vec b|^2}{|\vec a\times\vec b|^2}=\frac{16}{9}$$
So,$|\vec c|=\frac{4}{3}$
