Find the distance of the point with position vector ` -hat i-5 hat j-10 hat k` from the point of intersection of the line ` vec r=(2 hat i- hat j+2 hat k)+lambda(3 hat i+4 hat j+12 hat k)` with the plane ` vec rdot(( hat i- hat j+ hat k)=5.`

Explanation

Line:-`\{hat{r}} =(2 \hat{i}-\hat{j}+2 \hat{k})+\lambda(3 \hat{i}+4 \hat{j}+12 \hat{k})=(2+3 \lambda) \hat{i}+(-1+4 \lambda) \hat{j}+(2+12 \lambda) \hat{k}`
`(1)`


Plane:- `\hat{r}.(\hat{i}-\hat{j}+\hat{k})=5`
`(2)`


put `(1)` in `(2)`,


`{[(2+3 \lambda) \hat{i}+(-1+4 \lambda) \hat{j}+(2+12 \lambda) \hat{k}] \cdot(\hat{i}-\hat{j}+\hat{k})=5}`


`2+3 \lambda+1-4 \lambda+2+12 \lambda=5`


`11\lambda=0` ;`\lambda=0`


`\Rightarrow \hat{r}=2 \hat{i}-\hat{j}+2 \hat{k}; \vec{r}_{4}=-\hat{i}-5 \hat{j}-10 \hat{k}`


`=| \vec{r}-\vec{r}_{4}| =\sqrt{3^{2}+(-4)^{2}+12^{2}}=\sqrt{9+16+144}`


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