Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the plane passing through the intersection of the given planes is

$\vec{r}\cdot [(\hat{i}-2\hat{j}+3\hat{k})+\lambda (2\hat{i}+\hat{j}-3\hat{k})]=4+7\lambda$

$\vec{r}\cdot [(1+2\lambda )\hat{i}+(-2+\lambda )\hat{j}+(3-3\lambda )\hat{k}]=4+7\lambda$

Since this plane is parallel to the Y-axis, its normal vector is perpendicular to the Y-axis. The direction vector of the Y-axis is $\hat{j}$.

Taking the dot product of the normal vector with $\hat{j}$ and equating it to zero:

$(-2+\lambda )=0$

$\lambda =2$

Substituting $\lambda =2$ in the equation of the plane, we get:

$\vec{r}\cdot [(1+4)\hat{i}+(-2+2)\hat{j}+(3-6)\hat{k}]=4+14$

$\vec{r}\cdot (5\hat{i}-3\hat{k})=18$

Answer: $\vec{r}\cdot (5\hat{i}-3\hat{k})=18$