Equation of the plane that contains the line $\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-1}{1}$ and perpendicular to the plane 3x - y - z + 7 = 0 is:

x - 4y + 7z - 12 = 0

$\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-1}{1}.$

$\text{Direction ratios of line }=(1,2,1).$

$\text{Plane }3x-y-z+7=0 \Rightarrow \text{normal }(3,-1,-1).$

$\text{Required plane contains the line, so its normal is perpendicular to }(1,2,1).$

$\text{Also required plane is perpendicular to }3x-y-z+7=0.$

$\text{Hence its normal is perpendicular to }(3,-1,-1).$

$\text{Therefore normal }=\;(1,2,1)\times(3,-1,-1).$

$=\begin{vmatrix} i & j & k \\ 1 & 2 & 1 \\ 3 & -1 & -1 \end{vmatrix}.$

$=(-1)i+4j-7k.$

$\text{Normal }=(-1,4,-7).$

$\text{Point on line }(1,-1,1).$

$-1(x-1)+4(y+1)-7(z-1)=0.$

$-x+1+4y+4-7z+7=0.$

$x-4y+7z-12=0.$

$\text{Required plane }=x-4y+7z-12=0.$