Equation of the plane passing through the point $(2,0,5)$ and parallel to the vectors…
Equation of the plane passing through the point $(2,0,5)$ and parallel to the vectors $\hat{i}-\hat{j}+\hat{k}$ and $3 \hat{i}+2 \hat{j}+\hat{k}$ is
$x-4 y-z+3=0$
$x+4 y+5 z-27=0$
$x-4 y-5 z+23=0$
$x-4 y+z-7=0$
Solution
Normal to the plane is perpendicular to the given vectors.
Hence equation of normal is
$\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & -1 & 1 \\
3 & 2 & -1
\end{array}\right|=\hat{\mathrm{i}}(-1)-\hat{\mathrm{j}}(-4)+\hat{\mathrm{k}}(5)=-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$
Hence equation of required plane is
$\begin{aligned}
& (-1)(\mathrm{x}-2)+(4)(\mathrm{y}-0)+(5)(\mathrm{z}-5)=0 \\
& \therefore-\mathrm{x}+2+4 \mathrm{y}+5 \mathrm{z}-25=0 \Rightarrow \mathrm{x}-4 \mathrm{y}-5 \mathrm{z}+23=0
\end{aligned}$