Q.The equation of the plane in which the line $\dfrac{x-5}{4}=\dfrac{y-7}{4}=\dfrac{z+3}{-5}$ and $\dfrac{x-8}{7}=\dfrac{y-4}{1}=\dfrac{z+5}{3}$ lie, is
(A) 17x - 47y - 24z + 172 = 0
(B) 17x + 47y - 24z + 172 = 0
(C) 17x + 47y + 24z + 172 = 0
(D) 17x - 47y + 24z + 172 = 0

[!TLDR] Normal $=$ cross product of the two lines' direction ratios; plane through a point on either line. …

Line 1 passes through $(5,7,-3)$ with direction ratios $(4,4,-5)$. Line 2 passes through $(8,4,-5)$ with direction ratios $(7,1,3)$.

The plane containing both lines has normal equal to the cross product of the two direction-ratio triples:

$\vec n=\begin{vmatrix}\hat i&\hat j&\hat k\\4&4&-5\\7&1&3\end{vmatrix}=\hat i\big(4(3)-(-5)(1)\big)-\hat j\big(4(3)-(-5)(7)\big)+\hat k\big(4(1)-4(7)\big)$ …

Cross the two lines' direction ratios to get the common plane's normal, then use the point-normal form …

Sign errors in the determinant expansion for the normal (particularly the middle, $\hat j$, term which needs an extra negative sign); using the wrong point (f …